# Functonal Analysis 7: Further Properties of Normed Spaces

Normed spaces are not necessarily finite dimensional. So it is important to understand the notion of a basis for an infinite dimensional normed space. Suppose that there is a basis of a normed space $X$ as an infinite sequence $(e_n)$ in $X$. Then any $x\in X$ can be represented as the infinite superposition of the $e_n$’s

\label{eq:superposition}
x=\sum_{j=1}^\infty\alpha_je_j,

where $\alpha_1,\alpha_2,\cdots$ are scalars.
In order for this to make sense, we need to make sure that the infinite sum in \eqref{eq:superposition} converges. Thus we have the following definition of a basis for an infinite dimensional normed space.

Definition. Suppose that a normed space $X$ contains a sequence $(e_n)$ with property that $\forall x\in X$, there exists uniquely a sequence of scalars $(\alpha_n)$ such that
$$||x-\sum_{j=1}^n\alpha_je_j||\rightarrow 0\ \mbox{as}\ n\to\infty.$$
Then $(e_n)$ is called a Schauder basis for $X$. The infinite sum $\displaystyle\sum_{j=1}^\infty\alpha_je_j$ is called the expansion of $x$.

Example. $\ell^p$ has a Schauder basis $(e_n)$, where $e_n=(\delta_{nj})$.

Theorem. If a normed space has a Schauder basis, it is separable i.e. it has a countable dense subset.

Proof. Recall that $D$ is a dense subset of $X$ if $\bar D=X$. This equivalent to saying that $\forall \epsilon>0$, $\forall x\in X$, $B(x,\epsilon)\cap D\ne\emptyset$.

Let $D$ be the set of all possible finite linear combinations (superpositions) of the $e_n$’s. Then $D$ is countable. Let $x\in X$. Then $\exists$ a sequence of scalars $(\alpha_n)$ such that $x=\displaystyle\sum_{j=1}^\infty\alpha_je_j$. Given $\epsilon>0$, $\exists$ a positive integer $N$ such that
$$||x-(\alpha_1e_1+\cdots+\alpha_ne_n)||<\epsilon$$
for all $n\geq N$. This implies that $\alpha_1e_1+\cdots+\alpha_ne_n\in B(x,\epsilon)\cap D$ for all $n\geq N$.

One question mindful readers may have is does every separable Banach space have a Schauder basis? The answer is negative and a counterexample can be found in

Enflo, P. (1973), A counterexample to the approximation property. Acta Math. 130, 309–317.

We finish this lecture with the following theorem.

Theorem. [Completion] Let $X$ be a normed space. Then there exists a Banach space $\hat X$ and an isometry from $X$ onto $W\subset\hat X$ which is dense in $\hat X$. The space $\hat X$ is unique up to isometries.

# Factor Groups (Quotient Groups)

Let $G$ be a group and $N\leq G$. Recall the equivalence relation $\sim$ on $G$ defined by
$$a\sim b\ \mbox{if}\ ab^{-1}\in N$$ for any $a,b\in G$. Each equivalence class $[a]$ is identified with right coset $Na$. The quotient set $G/\sim$ is the set of all equivalence classes or equivalently all right cosets of the subgroup $N$ in $G$. A question we can ask is can we give a group structure to the quotient set $G/\sim$? The answer is affirmative if $N$ is a normal subgroup of $G$. This is what we are going to discuss in this lecture.

Let $N\vartriangleleft G$. Denote by $G/N$ the set of all right cosets of $N$ in $G$ i.e.
$$G/N=\{Na:a\in G\}.$$
For any $Na, Nb\in G/N$,
\begin{align*}
(Na)(Nb)&=N(aN)b\\
&=N(Na)b\\
&=N(Nab)\\
&=Nab.
\end{align*}
So, it appears that if $N\vartriangleleft G$, we may define an operation $\cdot$ on $G/N$ naturally from the binary operation on $G$ by the equation
\label{eq:cosetoper}
Na\cdot Nb=Nab
for any $Na,Nb\in G/N$. But is this operation well-defined? To see that let $Na=Nc$ and $Nb=Nd$. Then $Nac^{-1}=N$ and $Nbd^{-1}=N$. So $ac^{-1}\in N$ and $bd^{-1}\in N$. Now,
\begin{align*}
N(ac)(bd)^{-1}&=N(ac)(d^{-1}b^{-1})\\
&=Na(cd^{-1})b^{-1}\\
&=a(Ncd^{-1})b^{-1}\\
&=aNb^{-1}\\
&=Nab^{-1}\\
&=N.
\end{align*}
That is, $Na\cdot Nb=Nab=Ncd=Nc\cdot Nd$. Hence, right coset operation given by the equation \eqref{eq:cosetoper} is well-defined. Conversely, if $N\leq G$ and right coset operation given by the equation \eqref{eq:cosetoper} is well-defined, then $N$ must be a normal subgroup of $G$. The verification of this is left to readers as an exercise. Furthermore, $Ne=N$ is an identity element in $G/N$ and for each $Na\in G/N$, $Na^{-1}\in G$ and $NaNa^{-1}=N$, hence $(Na)^{-1}=Na^{-1}$. Therefore, $(G/N,\cdot)$ is a group. This group is called a factor group or a quotient group of $G$ modulo $N$.

Theorem. If $N\vartriangleleft G$ then there exists an onto homomorphism (an epimorphism) $\psi:G\longrightarrow G/N$ such that $\ker\psi=N$, Such a homomorphism is called a natural homomorphism or a canonical homomorphism.

Proof. Define $\psi: G\longrightarrow G/N$ by
$$\psi(a)=Na$$
for any $a\in G$. If $a=b$ then $\psi(a)=Na=Nb=\psi(b)$ so $\psi$ is well-defined. Let $Na\in G/N$. Then $a\in G$ and $\psi(a)=Na$, so $\psi$ is onto. $\psi(ab)=Nab=NaNb=\psi(a)\psi(b)$, hence $\psi$ is a homomorphism. To show that $\ker\psi=N$,
\begin{align*}
a\in\ker\psi&\Longleftrightarrow \psi(a)=N\\
&\Longleftrightarrow Na=N\\
&\Longleftrightarrow a\in N.
\end{align*}

$G/N$ is the set of all right coset of $N$ in $G$, so $|G/N|=|G:N|=\frac{|G|}{|N|}$.

Example. Since $\mathbb{Z}$ is an abelian group, for any $n\in\mathbb{N}$, $n\mathbb{Z}$ is a normal subgroup of $\mathbb{Z}$. So, for any $n\in\mathbb{N}$, $\mathbb{Z}/n\mathbb{Z}$ is a factor group. Each $\mathbb{Z}/n\mathbb{Z}$ is indeed isomorphic to $\mathbb{Z}_n$.

The notion of factor groups can be used to prove an important theorem in group theory called Cauchy Theorem. First we study the following theorem as we will also need it (actually its corollary) to prove Cauchy Theorem.

Theorem. Let $G$ be a finite group and $a\in G$ with $|a|=n$. Then for any $k\in\mathbb{Z}$,

1.  $|a^k|=\frac{n}{d}$ where $d=(k,n)$.
2.  $|a^k|=n$ if and only if $(k,n)=1$.

Proof. Since 2 follows from 1, we prove 1 only. Since $d=(k,n)$, $n=n_1d$ and $k=k_1d$ for some $n_1,k_1\in\mathbb{Z}$ such that $(n_1,k_1)=1$.
$$(a^k)^{n_1}=a^{k_1n}=(a^n)^{k_1}=e.$$ So, if we let $|a^k|=s$, then $s|n_1$. On the other hand, $e=|a^k|^s=a^{ks}$, so $n|ks$ which implies that $n_1|k_1s$. Since $(n-1,k_1)=1$, $n_1|s$. Therefore, $s=n_1=\frac{n}{d}$.

Corollary. If a finite group $G$ has no nontrivial subgroup, then $G$ must be a cyclic group of a prime order.

Proof. Let $a\ne e\in G$. Then $\langle a\rangle\leq G$. Since $G$ has no nontrivial subgroup and $a\ne e$, $\langle a\rangle=G$ i.e. $G$ is a cyclic group. Let $|a|=n$. If $k\ne n$ and $k|n$. Then by the above theorem $|a^k|=\frac{n}{k}$ since $(n,k)=k$. Since $a^k\ne e$ and $G$ has no nontrivial subgroup, $\langle a^k\rangle=G$. This implies that $|a^k|=n$ and so $k=1$. Therefore, $n$ is a prime.

Theorem. [Cauchy] If $G$ a finite abelian group and and $p||G|$ where $p$ is a prime, then $G$ has an element of order $p$.

Proof. We prove the theorem by induction on $|G|$. If $|G|=1$, then there is no prime that divides $|G|$, so the theorem is vacuously true. Suppose that the statement is true for all abelian groups whose order is less than $|G|$. Let $\{e\}\not\leq N\not\leq G$. If $p||N|$, then by induction hypothesis there exists $a\in N$ such that $|a|=p$. Since $N\subset G$, we are done in this case. Now suppose that $p\not|N|$. Since $G$ is abelian, $N\vartriangleleft G$. Since $p|G$ and $p\not| N$, $p||G/N|<|G|$. So by induction hypothesis, $G/N$ has an element $Na$ of order $p$.
$$(Na)^p=Na^p=N\Longrightarrow a^p\in N$$
but $a\not\in N$ since $Na\ne N$. Let $|N|=m$. Then $e=(a^p)^m=(a^m)^p$. Let $b=a^m$. If $b\ne e$, then $b$ is an element of order $p$. What if $b=a^m=e$? If so, $(Na)^m=N$. Since $|Na|=p$, $p|m=|N|$. but by assumption $p\not||N|$. A contradiction! So, we are done if $G$ has a nontrivial subgroup. What if $G$ does not have a nontrivial subgroup? If so, by corollary above $G$ must be a cyclic group of a prime order. Since $p||G|$, $|G|=p$. In this case, any $a\ne e\in G$ is an element of order $p$.

Remark. It should be noted that Cauchy Theorem still holds even if the group $G$ is not abelian. We will not however prove the generalized version of Cauchy Theorem here.

# Group Theory 9: Normal Subgroups

In here, it was shown that the kernel $K=\ker\varphi$ of a homomorphism $\varphi: G\longrightarrow G’$ is a subgroup of $G$ and that it satisfies the property $aK=Ka$ for all $a\in G$. The kernel of a homomorphism is an example of a particular kind of subgroups of a group called normal subgroups.

Definition. The subgroup $N$ of $G$ is called a normal subgroup of $G$ if $a^{-1}Na\subset N$, $\forall a\in G$. If $N$ is a normal subgroup of $G$, we write $N\vartriangleleft G$.

Note that $\forall a\in G$, $a^{-1}Na\subset N$ implies that $\forall a\in G$, $a^{-1}Na=N$. Hence, we have

Theorem. $N\vartriangleleft G$ if and only if $\forall a\in G$, $aN=Na$.

If $G$ is abelian, every subgroup is a normal subgroup. But the converse need not be true. There are nonabelian groups in which every subgroup is normal. Such nonabelian groups are called Hamiltonian.

Example. Let $\mathbb{Q}_8=\{\pm 1,\pm, i, \pm j, \pm k\}$ where $1,i,j,k$ satisfy the multiplication laws
\begin{align*}
1^2&=1,\ i^2=j^2=k^2=-1,\\
ij&=-ji=k,\ jk=-kj=i,\ ki=-ik=j.
\end{align*}
Then $\mathbb{Q}_8$ forms a nonabelian group. There are four subgroups of $\mathbb{Q}_8$:
\begin{align*}
\langle -1\rangle&=\{1,-1\},\\
\langle i\rangle&=\{1,-1,i,-i\},\\
\langle j\rangle&=\{1,-1,j,-j\},\\
\langle k\rangle&=\{1,-1,k,-k\}.
\end{align*}
All these subgroups are normal subgroups of $\mathbb{Q}_8$, so $\mathbb{Q}_8$ is Hamiltonian. Note that $1,i,j,k$ form a 4-dimensional algebra of quaternions over $\mathbb{R}$ $$\mathbb{H}=\{a1+bi+cj+dk:a,b,c,d\in\mathbb{R}\}.$$ The real algebra of quaternions $\mathbb{H}$ is identified with the 4-dimensional Euclidean space $\mathbb{R}^4$.

Recall that the index $|G:H|$ of a subgroup $H$ in $G$ is the number of right cosets of $H$ in $G$ or equivalently the number of left cosets of $H$ in $G$ as discussed here.

Theorem. Let $G$ be a group and $N\leq G$. If $|G:N|=2$ then $N\vartriangleleft G$.

Proof. Let $a\in G$. If $a\in N$, $aN=Na$. Now suppose that $a\in G\setminus N$. Then
$$N\cap Na=\emptyset=N\cap aN.$$
Furthermore, since $G=N\cup Na=N\cup aN$, $Na=Na$.

Example. Recall that $|S_n|=n!$ and $|A_n|=\frac{n!}{2}$. Since $|S_n:A_n|=2$, $A_n\vartriangleleft S_n$.

Remark. The converse of the above theorem need not be true, namely the index $|G:N|$ of a normal subgroup $N$ in $G$ is not necessarily 2. The 4th dihedral group (we introduced it here)
$$D_4=\{1,\sigma,\sigma^2,\sigma^3,\tau,\sigma\tau,\sigma^2\tau,\sigma^3\tau\}$$
has the following nontrivial subgroups:
\begin{align*}
&\{1,\sigma^2,\sigma\tau,\sigma^2\tau\},\ \{1,\sigma,\sigma^2,\sigma^3\},\ \{1,\sigma^2,\tau,\sigma^2\tau\},\\
&\{1,\sigma^3\tau\},\ \{1,\sigma\tau\},\ \{1,\sigma^2\},\ \{1,\tau\},\ \{1,\sigma^2\tau\}.
\end{align*}
The subgroups of order 4 are normal subgroups of $D_4$ because their indices are 2. The center of $D_4$ is $\{1,\sigma^2\}$, so it is also a normal subgroup of $D_4$. However, $|D_4:Z(D_4)|=4$. The rest of nontrivial subgroups are not normal subgroups of $D_4$.

Remark. $H\vartriangleleft N, N\vartriangleleft G\not\Longrightarrow H\vartriangleleft G$. In the previous remark, $H=\{1,\tau\}$ is a normal subgroup of $\{1,\sigma^2,\tau,\sigma^2\tau\}$ but $H\not\vartriangleleft D_4$.

Example. The center $Z(G)$ of a group $G$
$$Z(G)=\{x\in G: \forall g\in G, xg=gx\}$$ is a normal subgroup of $G$.

Example. If $N\leq Z(G)$ then $N\vartriangleleft G$.

Example. Let $\varphi: G\longrightarrow G’$ be a homomorphism. Then $\ker\varphi\vartriangleleft G$ as seen here.

Example. Let $\mathbb{R}^\ast=\mathbb{R}\setminus\{0\}$. Define $f: \mathrm{GL}(2,\mathbb{R})\longrightarrow\mathbb{R}^\ast$ by
$$f(A)=\det A,\ \mbox{for any}\ A\in\mathrm{GL}(2,\mathbb{R}).$$ Then $f$ is a homomorphism because for any $A,B\in\mathrm{GL}(2,\mathbb{R})$, $f(AB)=\det(AB)=(\det A)(\det B)=f(A)f(B)$. $f$ is also onto: for any $r\in\mathbb{R}^\ast$ let $A=\begin{pmatrix} r & 0\\ 0 & 1 \end{pmatrix}$. Then $A\in\mathrm{GL}(2,\mathbb{R})$ and $f(A)=r$. For any $A\in\mathrm{GL}(2,\mathbb{R})$, $A\in\ker f$ if and only if $\det A=1$ and so $\ker f=\mathrm{SL}(2,\mathbb{R})$. Hence, $\mathrm{SL}(2,\mathbb{R})\vartriangleleft\mathrm{GL}(2,\mathbb{R})$.

# Functional Analysis 6: Normed Spaces and Banach Spaces

From here on, I assume a background of undergraduate level Linear Algebra. Readers should be familiar with notions such as vector spaces, subspaces, linear combination, linear dependence and linear independence, basis, and dimension.

In this lecture, we begin with the notion of a normed space. A normed space is a vector space with a norm defined. So what is a norm? A norm $||\cdot||$ is a real-valued function $||\cdot||: V\longrightarrow\mathbb{R}^+\cup\{0\}$ such that for $x,y\in V$ and for $\alpha$ a scalar,

(N1) $||x||=0\Longleftrightarrow x=O$.

(N2) $||\alpha x||=|\alpha|||x||$.

(N3) $||x+y||\leq ||x||+||y||$. (Triangle Inequality.)

A norm on $X$ defines a metric $d$ on $X$

\label{eq:metric}
d(x,y)=||x-y||.

\eqref{eq:metric} is called a metric induced by the norm $||\cdot||$. So, a normed space is a metric space but the converse need not be true.

A complete normed space is called a Banach space.

Example. $\mathbb{R}^n$ and $\mathbb{C}^n$ with the Euclidean norm
$$||x||=\left(\sum_{j=1}^n|\xi_j|^2\right)^{\frac{1}{2}}$$ are Banach spaces.

Example. $\ell^p$ with the norm
$$||x||=\left(\sum_{j=1}^n|\xi_j|^p\right)^{\frac{1}{p}}$$ is a Banach space.

Example. $\ell^\infty$ with the norm
$$||x||=\sup_{j\in\mathbb{N}}|\xi_j|$$ is a Banach space.

Example. $\mathcal{C}[a,b]$ with the norm
$$||x||=\max_{t\in[a,b]}|x(t)|$$ is a Banach space.

What follows next is an example of a normed space which is not complete. $\mathcal{C}[a,b]$, the vector space of all continuous real-valued functions on $[a,b]$ forms a normed space with the norm defined by

\label{eq:intnorm}
||x||=\left(\int_a^b x(t)^2dt\right)^{\frac{1}{2}}.

Let $[a,b]=[0,1]$. for each $n=1,2,\cdots$, let $a_n=\frac{1}{2}+\frac{1}{n}$. Define a sequence $(x_n)$ in $\mathcal{C}[0,1]$ by
$$x_n(t)=\left\{\begin{array}{ccc} 0 & \mbox{if} & t\in\left[0,\frac{1}{2}\right],\\ nt-\frac{n}{2} & \mbox{if} & t\in\left[\frac{1}{2},a_n\right],\\ 1 & \mbox{if} & t\in[a_n,1]. \end{array}\right.$$

Let us assume that $n>m$. Then
\begin{align*}
||x_m-x_n||^2&=\int_0^1(x_m(t)-x_n(t))^2dt\\
&=\int_0^{\frac{1}{2}}(x_m(t)-x_n(t))^2dt+\int_{\frac{1}{2}}^{a_n}(x_m(t)-x_n(t))^2dt\\&+\int_{a_n}^{a_m}(x_m(t)-x_n(t))^2dt+\int_{a_m}^1(x_m(t)-x_n(t))^2dt\\
&=\int_{\frac{1}{2}}^{\frac{1}{2}+\frac{1}{n}}\left(nt-\frac{n}{2}-mt+\frac{m}{2}\right)^2dt+\int_{\frac{1}{2}+\frac{1}{n}}^{\frac{1}{2}+\frac{1}{m}}\left(1-mt+\frac{m}{2}\right)^2dt\\
&=\frac{(n-m)^2}{3mn^2}\\
&<\frac{1}{3m}-\frac{1}{3n}.
\end{align*}
Given $\epsilon>0$, choose a positive integer $N$ so that $N>\frac{1}{3\epsilon^2}$. Then for all $m,n\geq N$,
\begin{align*}
||x_n-x_m||^2&<\frac{1}{3m}-\frac{1}{3n}\\
&<\frac{1}{3m}\\
&\leq\frac{1}{3N}\\
&<\epsilon^2.
\end{align*}
Therefore, $(x_n)$ is a Cauchy sequence in $\mathcal{C}[0,1]$. For any $x(t)\in\mathcal{C}[0,1]$,
\begin{align*}
||x_n-x||^2&=\int_0^1(x_n(t)-x(t))^2dt\\
&=\int_0^{\frac{1}{2}}x(t)^2dt+\int_{\frac{1}{2}}^{a_m}(x_n(t)-x(t))^2dt+\int_{a_n}^1(1-x(t))^2dt.
\end{align*}
Suppose that $x_n\rightarrow x$ as $n\to\infty$. The by the continuity of $x_n(t)$ and $x(t)$, $x(t)$ must satisfy that
$$x(t)=\left\{\begin{array}{ccc} 0 & \mbox{if} & t\in\left[0,\frac{1}{2}\right),\\ 1 & \mbox{if} & t\in\left(\frac{1}{2},1\right]. \end{array}\right.$$
But this is impossible since $x(t)$ is continuous. Hence, $(x_n)$ is not convergent in $\mathcal{C}[0,1]$.

In here, we studied completion of metric spaces. Since a normed space is a metric space, we also have completion of normed spaces. The completion of $\mathcal{C}[a,b]$ with the norm \eqref{eq:intnorm} is denoted by $L^2[a,b]$. More generally, for any $p\geq 1$, the Banach space $L^p[a,b]$ is the completion of the normed space $\mathcal{C}[a,b]$ with the norm

||x||_p=\left(\int_a^b|x(t)|^pdt\right)^{\frac{1}{p}}.

Lemma. A metric $d$ induced by a norm on a normed space $X$ satisfies

1. $d(x+a,y+a)=d(x,y)$, $a,x,y\in X$. This means that $d$ is translation invariant.
2. $d(\alpha x,\alpha y)=|\alpha|d(x,y)$, $\alpha$, a scalar.

Proof.
\begin{align*}
d(x+a,y+a)&=||x+a-(y+a)||=||x-y||=d(x,y),\\
d(\alpha x,\alpha y)&=||\alpha x-\alpha y||=|\alpha|||x-y||=|\alpha|d(x,y).
\end{align*}

We know that a norm on a vector space $V$ defines a metric on $V$. Can every metric on a vector space $V$ be obtained from a norm? The answer is negative. Let $V$ be the set of all bounded or unbounded sequences of complex numbers. Then $V$ is a vector space. The metric $d$ on $V$ defined by
$$d(x,y)=\sum_{j=1}^\infty\frac{1}{2^j}\frac{|\xi_j-\eta_j|}{1+|\xi_j-\eta_j|}$$ is not translation invariant, so it cannot be obtained from a norm.

# Group Theory 8: Homomorphisms

In this lecture, we study maps from a group into another group that carry over the same group structure i.e preserve group operations. Such maps are called homomorphisms. It turns out that it suffices to require a homomorphism to preserve a binary operation only. Then the unary operation (inverses) and the nullary operation (the identity element) are automatically preserved.

Definition. Let $(G,\ast)$ and $(G’,\sharp)$ be two groups. A map $\varphi: G\longrightarrow G’$ is called a homomorphism if $\varphi(a\ast b)=\varphi(a)\sharp\varphi(b)$.

Example. Define $\varphi: (\mathbb{R}^+,\cdot)\longrightarrow(\mathbb{R},+)$ by
$$\varphi(x)=\log_{10}x,\ \forall x\in\mathbb{R}^+.$$
Then $\varphi$ is a homomorphism. In addition, $\varphi$ is one-to-one and onto. A homomorphism which is one-to-one and onto is called an isomorphism. If there is an isomorphism from a group onto another group, they are the same group meaning we do not distinguish them.

Example. The map $\varphi: \mathbb{Z}\longrightarrow\mathbb{Z}_n$ defined by
$$\varphi(x)=[x],\ \forall\ x\in\mathbb{Z}$$
is a homomorphism called the canonical or the natural homomorphism. The canonical homomorphism is onto.

Example. Let $G$ be a group and $A(G)$ the set of all bijective (i.e. one-to-one and onto) maps of $G$ onto itself. Recall that $A(G)$ forms a group with composition. Let $a\in G$. Define $T_a: G\longrightarrow G$ by
$$T_ax=ax\ \forall\ x\in G.$$ Hence $\forall a\in G$, $T_a\in A(G)$. Furthermore, $\forall a,b\in G$, $T_aT_b=T_{ab}$: $\forall x\in G$,
$$T_aT_b(x)=T_a(T_bx)=T_a(bx)=a(bx)=(ab)x=T_{ab}x.$$
Define $\varphi: G\longrightarrow A(G)$ by
$$\varphi(a)=T_a,\ \forall\ a\in G.$$ Then $\varphi$ is a homomorphism. In addition, it is one-to-one. If $|G|=n$, then $|A(G)|=n!$, so $\varphi$ cannot be onto.

Definition. A homomorphism which is one-to-one (injective) is called a monomorphism. A homomorphism which is onto (surjective) is called an epimorphism. A homomorphism which is one-to-one and onto (bijective) is called an isomorphism as mentioned above. $G$ is said to be isomorphic to $G’$ if there exists an isomorphism from $G$ onto $G’$. If $G$ is isomorphic to $G’$, we write $G\cong G’$. An isomorphism from a group $G$ onto itself is called an automorphism.

Theorem. [Cayley's Theorem] Every group $G$ is isomorphic to some subgroup of $A(S)$, for an appropriate set $S$.

We in fact proved this theorem in the previous example by taking $S$ to be the group $G$ itself. But there may be other choices for $S$. If $G$ is finite, $S$ may be taken to be a finite set in which case $A(S)$ is $S_n$, the group of permutations of $\{1,2,\cdots,n\}$. In this case, Cayley’s Theorem is stated as: A finite group can be represented as a group of permutations.

Example. Let $G$ be a group and $a\in G$ be fixed. The map $\varphi: G\longrightarrow G$ defined by
$$\varphi(x)=a^{-1}xa,\ \forall x\in G$$
is an automorphism called the inner automorphism of $G$ induced by $a$.

Example. Define $\varphi: (\mathbb{R},+)\longrightarrow (\mathbb{C}\setminus\{0\},\cdot)$ by
$$\varphi(x)=e^{ix},\ \forall x\in\mathbb{R}.$$
While $\varphi$ is a homomorphism, it is neither one-to-one nor onto.

Lemma. Let $\varphi: G\longrightarrow G’$ be a homomorphism. Then

(a) $\varphi(e)=e’$.

(b) $\varphi(a^{-1})=\varphi(a)^{-1}$.

Proof. (a) Let $a\in G$. Then $ae=a$. Since $\varphi$ is a homomorphism, $$\varphi(ae)=\varphi(a)\varphi(e)=\varphi(a)=\varphi(a)e’.$$ So by the cancellation law, we obatin $\varphi(e)=e’$.

(b) Let $a\in G$. Then
\begin{align*}
\varphi(a)\varphi(a^{-1})&=\varphi(aa^{-1})\ (\mbox{because $\varphi$ is a homomorphism})\\
&=\varphi(e)\\
&=e’\ (\mbox{by part (a)})
\end{align*}
Hence, $\varphi(a)^{-1}=\varphi(a^{-1})$.

Lemma. Let $\varphi: G\longrightarrow G’$ be a homomorphism. Then $\varphi(G)\leq G’$.

In here, the kernel of a function was introduced. Let $\varphi: G\longrightarrow G’$ be a homomorphism and $K=\varphi^{-1}(e’)=\{a\in G: \varphi(a)=e’\}$. Then $\ker\varphi$ and $K=\varphi^{-1}(e’)$, the preimage of $e’$ under $\varphi$ are closely related as:
\begin{align*}
(a,b)\in\ker\varphi&\Longleftrightarrow\varphi(a)=\varphi(b)\\
&\Longleftrightarrow\varphi(ab^{-1})=e’\\
&\Longleftrightarrow ab^{-1}\in K.
\end{align*}
So, in group theory we define $K=\varphi^{-1}(e’)$ to be $\ker\varphi$, the kernel of $\varphi$.

Theorem. $\varphi: G\longrightarrow G’$ is one-to-one if and only if $\ker\varphi=\{e\}$.

Proof. ($\Rightarrow$) Suppose that $\varphi$ is one-to-one and let $a\in\ker\varphi$. Then $\varphi(a)=\varphi(e)=e’$. Since $\varphi$ is one-to-one, $a=e$.

($\Leftarrow$) Suppose that $\ker\varphi=\{e\}$ and let $\varphi(a)=\varphi(b)$. Then
\begin{align*}
\varphi(ab^{-1})=e’&\Longrightarrow ab^{-1}\in\ker\varphi\\
&\Longrightarrow ab^{-1}=e\\
&\Longrightarrow a=b.
\end{align*}
So, $\varphi$ is one-to-one.

Theorem. Let $\varphi: G\longrightarrow G’$ be a homomorphism. Then $\ker\varphi\leq G$.

Theorem. Let $K=\ker\varphi$. Then $\forall a\in G$, $aK=Ka$.

Proof. It suffices to show (why?) that $\forall a\in G$, $a^{-1}Ka\subset K$. Let $a\in G$. Then $\forall k\in K$,
$$\varphi(a^{-1}ka)=\varphi(a)^{-1}\varphi(k)\varphi(a)=e’\Longrightarrow a^{-1}ka\in K.$$

# Group Theory 7: Congruence Modulo $n$

In this lecture, we study the congruence modulo $n$, $\equiv\mod n$ on $\mathbb{Z}$. As we shall see, $\equiv\mod n$ is an equivalence relation on $\mathbb{Z}$. It is indeed more than just an equivalence relation. $\equiv\mod n$ also preserves operations $+$ and $\cdot$ on $\mathbb{Z}$. Because of this, we can define an operation $+$ on the quotient set $\mathbb{Z}/\equiv\mod n$ so that it becomes a quotient group. In general, if an equivalence relation preserves operations, it is called a congruence relation and a congruence relation allows us to define a quotient algebra e.g. a quotient group. Let $G$ be a group and $H\leq G$. We studied the equivalence relation $\sim$ on $G$: $\forall a,b\in G$,
$$a\sim b \Longleftrightarrow ab^{-1}\in H.$$ It turns out that $\sim$ is a congruence relation i.e. it preserves the group operation $\cdot$ if $H$ is a normal subgroup of $G$. I am not going to talk about normal subgroups here but we will study details about this next time.

On $\mathbb{Z}$, define $a\equiv b\mod n$, we say $a$ is congruent to $b$ modulo $n$, if $n|b-a$. The class of $a$
$$[a]=\{a+nk: k\in\mathbb{Z}\}$$ is called the congruence class of $a$. By Euclid’s algorithm, $\forall b\in\mathbb{Z}$, $b=qn+r$ where $0\leq r<n$. This implies that $b\equiv r\mod n$and so $[b]=[r]$. Hence, the $n$ classes $[0],[1],\cdots,[n-1]$ give us all the congruence classes and they are all distinct. Let
$$\mathbb{Z}_n=\{[0],[1],\cdots,[n-1]\}.$$ That is, $\mathbb{Z}_n$ is the quotient set modulo the equivalence relation $\equiv\mod n$, $\mathbb{Z}/\equiv\mod n$. Define $+$ on $\mathbb{Z}_n$: $\forall [a],[b]\in\mathbb{Z}_n$,
$$[a]+[b]:=[a+b].$$ Then $+$ is well-defined on $\mathbb{Z}_n$. What this means is that $+$ is well-defined as a function $+:\mathbb{Z}_n\times\mathbb{Z}_n\longrightarrow\mathbb{Z}_n$. In other words, if $[a]=[a']$ and $[b]=[b']$ then $[a+b]=[a'+b']$. This can be shown straightforward and is left for the readers. One can also readily see that $(\mathbb{Z}_n,+)$ is an abelian group. Define $\cdot$ on $\mathbb{Z}_n$: $\forall [a],[b]\in\mathbb{Z}_n$,
$$[a]\cdot[b]:=[ab].$$ Then $\cdot$ is also well-defined on $\mathbb{Z}_n$. However, $(\mathbb{Z}_n,\cdot)$ is not a group. $\forall [a]\in\mathbb{Z}_n$, $[0][a]=[0]$ and $[1]$ is the identity under multiplication, so $[0]$ does not have a multiplicative inverse. Would nonzero elements of $\mathbb{Z}_n$ form a group then? The answer is no. For instance, in $\mathbb{Z}_6$, $[2],[3]\ne 0$ but $[2][3]=[6]=[0]$. So, is there a subset of $\mathbb{Z}_n$ that forms a group under $\cdot$? The answer is affirmative. Let $\mathbb{Z}_n^\ast=\{[a]\in\mathbb{Z}_n: (a,n)=1\}$. First, note that for $[a]=[b]$, $(a,n)=1$ if and only if $(b,n)=1$. Recall the property that if $(a,n)=1$ and $(b,n)=1$, then $(ab,n)=1$. This implies that $[a][b]=[ab]\in\mathbb{Z}_n^\ast$. Now suppose that $[a][b]=[a][c]$. Then
\begin{align*}
[ab]=[ac]&\Longrightarrow ab-ac=nk\\
&\Longrightarrow n|a(b-c)\\
&\Longrightarrow n|b-c\ (\mbox{since}\ (a,n)=1)\\
&\Longrightarrow [b]=[c].
\end{align*}
That is, $\mathbb{Z}_n^\ast$ has the cancellation property, so $(\mathbb{Z}_n^\ast,\cdot)$ is a group. In some textbooks, $\mathbb{Z}_n^\ast$ is also denoted by $U_n$. The order of $\mathbb{Z}_n^\ast$ is the number of integers $1\leq m<n$ such that $(m,n)=1$.

Definition. The Euler $\varphi$-function, $\varphi(n)$, is defined by $\varphi(1)=1$ and for $n>1$, $\varphi(n)=$ the number of positive integers $m$ with $1\leq m<n$ such that $(m,n)=1$. Hence, $|\mathbb{Z}_n^\ast|=\varphi(n)$. If $n=p$, a prime, then $\varphi(p)=p-1$.

Note that the Euler $\varphi$-function is multiplicative, namely if $(m,n)=1$, then $\varphi(mn)=\varphi(m)\varphi(n)$.

Example. $\varphi(8)=4$, $\varphi(15)=8$.

Example.
\begin{align*}
\mathbb{Z}_8^\ast&=\{[1],[3],[5],[7]\},\\
\mathbb{Z}_{15}^\ast&=\{[1],[2],[4],[7],[8],[11],[13],[14]\}.
\end{align*}

Example. $\mathbb{Z}_9^\ast=\{[1],[2],[4],[5],[7],[8]\}$.
\begin{align*}
[1]^1&=[2],\ [2]^2=[4],\ [2]^3=[8],\ [2]^4=[16]=[7],\\
[2]^5&=[32]=[5],\ [2]^6=[2][2]^5=[2][5]=[10]=[1].
\end{align*}
Hence, $\mathbb{Z}_9^\ast$ is a cyclic group $\mathbb{Z}_9^\ast=\langle[2]\rangle$.

As a summary,

Theorem. $\mathbb{Z}_n^\ast$ forms an abelian group under the product $[a][b]=[ab]$, of order $\varphi(n)$, where $\varphi(n)$ is the Euler $\varphi$-function.

Theorem. [Euler] If $(a,n)=1$ then $a^{\varphi(n)}\equiv 1\mod n$.

Proof. Since $\mathbb{Z}_n^\ast$ is a group of order $\varphi(n)$, $a^{\varphi(n)}=1$, $\forall a\in\mathbb{Z}_n^\ast$. So,
$$[a^{\varphi(n)}]=[a]^{\varphi(n)}=[1]\Longrightarrow a^{\varphi(n)}\equiv 1\mod n.$$

Corollary. [Fermat] If $p$ is a prime and $p\not | a$, then
$$a^{p-1}\equiv 1\mod p.$$
For any integer $b$, $b^p\equiv b\mod p$. This corollary is usually called Fermat’s Little Theorem.

Proof. $\varphi(p)=p-1$. If $(a,p)=1$, then by the previous theorem,
\begin{align*}
a^{p-1}\equiv 1\mod p &\Longrightarrow a^{p-1}-1=pk\\
&\Longrightarrow a^p-a=p(ak)\\
&\Longrightarrow a^p\equiv a\mod p.
\end{align*}
If $p\not| b$, then $b^p\equiv b\mod p$. If $p|b$, then $b\equiv 0\mod p\Longrightarrow b^p\equiv 0\mod p$. So for any $b\in\mathbb{Z}$, $b^p\equiv b\mod p$.

Example. Compute the remainder of $8^{103}$ when divided by 13.

Solution. Since $13\not|8$, by Fermat’s Little Theorem we have $8^{13-1}=8^{12}\equiv 1\mod 13$. Dividing 103 by 13, we obtain $103=12\cdot 8+7$. So,
\begin{align*}
8^{103}&\equiv (8^{12})^88^7\equiv 8^7\equiv (-5)^7\ (8\equiv -5\mod 13)\\
&\equiv (25)^3(-5)\equiv (-1)^3(-5)\equiv 5\mod 13\ (25\equiv -1\mod 13).
\end{align*}

Example. Show that $2^{11,213}-1$ is not divisible by 11.

Solution. Since $11\not| 2$, by Fermat’s Little Theorem we have $2^{10}\equiv 1\mod 11$. $11,213=10\cdot 1,121+3$. So,
\begin{align*}
2^{11,213}-1&\equiv (2^{10})^{1,121}\cdot 2^3-1\equiv 2^3-1\\
&\equiv 7\mod 11.
\end{align*}
Hence, when $2^{11,213}-1$ is divided by 11, the remainder is 7.

The number $11,213$ is a prime and $2^{11,213}-1$ is also a prime. In number theory, primes of the form $2^p-1$ where $p$ is a prime are called Mersenne primes.

# Functional Analysis 5: Completion of Metric Spaces

This lecture will conclude our discussion on metric spaces with completion of metric spaces.

Definition. Let $X=(X,d)$ and $\tilde X=(\tilde X,d)$ be two metric space. A mapping $T: X\longrightarrow\tilde X$ is said to be an isometry of $T$ preserves distances, i.e.
$$\forall x,y\in X,\ \tilde d(Tx,Ty)=d(x,y).$$ The space $X$ is said to be isometric with $\tilde X$ if there exists a bijective isometry of $X$ onto $\tilde X$.

Theorem [Completion] For a metric space $(X,d)$ there exists a complete metric space $\hat X=(\hat X,\hat d)$ which has s subspace $W$ that is isometric with $X$ and is dense in $\hat X$. $\hat X$ is called the completion of $X$ and it is unique up to isometries.

Proof. This will be a lengthy proof and I have divided it into steps.

Step 1. Construction of $\hat X=(\hat X,\hat d)$.

Let $(x_n)$ and $(y_n)$ be Cauchy sequences in $X$. We say $(x_n)$ is equivalent to $(x_n’)$, and write $(x_n)\sim (x_n’)$, if
$$\lim_{n\to\infty}d(x_n,x_n’)=0.$$ $\sim$ is actually an equivalence relation on the set of all Cauchy sequences of $X$. Clearly $\sim$ is reflexive and symmetric. Let us show that $\sim$ is also transitive. Let $(x_n)\sim (x_n’)$ and $(x_n’)\sim (x_n^{\prime\prime})$. Then
$$\lim_{n\to\infty}d(x_nx_n’)=0\ \mbox{and}\ \lim_{n\to\infty}d(x_n’,x_n^{\prime\prime})=0.$$
\begin{align*}
\lim_{n\to\infty}d(x_n,x_n^{\prime\prime})&\leq\lim_{n\to\infty}d(x_n,x_n’)+\lim_{n\to\infty}d(x_n’,x_n^{\prime\prime})\\
&=0
\end{align*}
Let $\hat X$ be the set of all equivalence classes $\hat x,\hat y,\cdots$ of Cauchy sequences. Define
$$\hat d(\hat x,\hat y)=\lim_{n\to\infty}d(x_n,y_n),$$
where $x_n\in\hat x$ and $y_n\in\hat y$. We claim that $\hat d$ is a metric on $\hat X$. It suffices to show that $\hat d$ is well-defined. The conditions (M1)-(M3) hold due to the fact that $d$ is a metric. First we show that $\hat d(\hat x,\hat y)$ exists. It follows from (M3) that
$$d(x_n,y_n)\leq d(x_n,x_m)+d(x_m,y_m)+d(y_m,y_n)$$ and so we have
$$d(x_n,y_n)-d(x_m,y_m)\leq d(x_n,x_m)+d(y_m,y_n).$$ Similarly, we obtain
$$d(x_m,y_m)-d(x_n,y_n)\leq d(x_n,x_m)+d(y_m,y_n).$$ Hence,
$$|d(x_n,y_n)-d(x_m,y_m)|\leq d(x_n,x_m)+d(y_m,y_n)\rightarrow 0$$ as $n,m\rightarrow\infty$, i.e.
$$\lim_{n,m\to\infty}|d(x_n,y_n)-d(x_m,y_m)|=0.$$
Since $\mathbb{R}$ is complete, $\displaystyle\lim_{n\to\infty}d(x_n,y_n)$ exists. Now we show that the limit is independent of the choice of representatives $(x_n)$ and $(y_n)$. If $(x_n)\sim (x_n’)$ and $(y_n)\sim(y_n’)$, then
$$|d(x_n,y_n)-d(x_n’,y_n’)|\leq d(x_n,x_m)+d(y_m,y_n)\rightarrow 0$$ as $n\to\infty$.

Step 2. Construction of an isometry $T: X\longrightarrow W\subset \hat X$.

For each $b\in X$, let $\hat b$ be the equivalence class of the Cauchy sequence $(b,b,b,\cdots)$. Then $T(b):=\hat b\in\hat X$. Now,
$$\hat d(Tb,Tc)=\hat d(\hat b,\hat c)=d(b,c).$$ So, $T$ is an isometry. An isometry is automatically injective. $T$ is onto since since $T(X)=W$. Let us show that $W$ is dense in $\hat X$. Let $\hat x\in \hat X$ and let $(x_n)\in\hat x$. Since $(x_n)$ is Cauchy, given $\epsilon>0$ $\exists N$ such that $d(x_n,x_N)<\frac{\epsilon}{2}$ $\forall n\geq N$. Let $(x_N,x_N,\cdots)\in\hat x_N$. Then $\hat x_N\in W$.
\begin{align*}
\hat d(\hat x,\hat x_N)=\lim_{n\to\infty}d(x_n,x_N)\leq\frac{\epsilon}{2}<\epsilon&\Longrightarrow \hat x_N\in B(\hat x,\epsilon)\\
&\Longrightarrow B(\hat x,\epsilon)\cap W\ne\emptyset.
\end{align*}
Hence, $\bar W=\hat X$ i.e. $W$ is dense in $\hat X$.

Step 3. Completeness of $\hat X$.

Let $(\hat x_n)$ be any Cauchy sequence in $\hat X$. Since $W$ is dense in $\hat X$, $\forall \hat x_n$, $\exists\hat z_n\in W$ such that $\hat d(\hat x_n,\hat z_n)<\frac{1}{n}$.
\begin{align*}
\hat d(\hat z_m,\hat z_n)&\leq \hat d(\hat z_n,\hat x_m)+\hat d(\hat x_m,\hat x_n)+\hat d(\hat x_n,\hat z_n)\\
&<\frac{1}{m}+\hat d(\hat x_m,\hat x_n)+\frac{1}{n}.
\end{align*}
Given $\epsilon>0$ by Archimedean property $\exists$ a positive integer $N_1$ such that $N>\frac{\epsilon}{3}$. Since $(\hat x_n)$ is a Cauchy sequence, $\exists$ a positive integer $N_2$ such that $\hat d(\hat x_m,\hat x_n)<\frac{\epsilon}{3}$ $\forall m,n\geq N$. Let $N=\max\{N_1,N_2\}$. Then $\forall m,n\geq N$, $\hat d(\hat z_m,\hat z_n)<\epsilon$ i.e. $(\hat z_m)$ is Cauchy. Since $T: X\longrightarrow W$ is an isometry and $\hat z_m\in W$, the sequence $(z_m)$, where $z_m=T^{-1}\hat z_m$, is Cauchy in $X$. Let $\hat x\in\hat X$ be the class to which $(z_m)$ belongs. Show that $\hat x$ is the limit of $(\hat x_n)$. For each $n=1,2,\cdots$,
\begin{align*}
\hat d(\hat x_n,\hat x)&\leq \hat d(\hat x_n,\hat z_n)+\hat d(\hat z_n,\hat x)\\
&<\frac{1}{n}+\hat d(\hat z_n,\hat x)\\
&=\frac{1}{n}+\lim_{m\to\infty}d(z_n,z_m)
\end{align*} since $(z_m)\in\hat x$ and $(z_n,z_n,\cdots)\in\hat z_n\in W$. This implies that $\displaystyle\lim_{n\to\infty}\hat d(\hat x_n,\hat x)=0$ i.e. the Cauchy sequence $(\hat x_n)$ in $\hat X$ has the limit $\hat x\in\hat X$. Therefore, $\hat X$ is complete.

Step 4. Uniqueness of $\hat X$ up to isometries.

Suppose that $(\tilde X,\tilde d)$ is another completion of $X$ i.e. it is a complete metric space with a subspace $\tilde W$ dense in $\tilde X$ and isometric with $X$. We show that $\hat X$ is isometric with $\tilde X$. Let $X$ is isometric with $W$ and $\tilde W$ via isometries $T$ and $\tilde T$, respectively. Then $W$ is isometric with $\tilde W$ via the isometry $\rho=\tilde T\circ T^{-1}$.

$$\begin{array}{ccc} & & W\\ & \nearrow &\downarrow\\ X & \longrightarrow & \tilde{W} \end{array}$$

Let $\hat x\in\hat X$. Then $\exists$ a sequence in $(\hat x_n)\in W$ such that $\displaystyle\lim_{n\to\infty}\hat x_n=\hat x$. $(\hat x_n)$ is a Cauchy sequence and $\rho$ is an isometry, so $(\tilde x_n)$, where $\tilde x_n:=\rho\hat x_n$, is a Cauchy sequence in $\tilde W\subset \tilde X$. Since $\tilde X$ is complete, $\exists\tilde x\in\tilde X$ such that $\displaystyle\lim_{n\to\infty}\tilde x_n=\tilde x$. Define a mapping $\psi:\hat X\longrightarrow\tilde X$ by $\psi\hat x=\tilde x$. Then we claim that $\hat X$ is isometric with $\tilde X$ via $\psi$.

Step A. $\psi$ is well-defined.

It suffices to show that $T\hat x$ does not depend on the choice of $(\hat x_n)\in W$ such that $\displaystyle\lim_{n\to\infty}\hat x_n=\hat x$. Let $(\hat x_n’)$ be another sequence in $W$ such that $\displaystyle\lim_{n\to\infty}\hat x_n’=\hat x$. Then $(\tilde x_n’)$, where $\tilde x_n’=\rho\hat x_n’$, is a Cauchy sequence in $\tilde W$ and so $\exists\tilde x’\in\tilde X$ such that $\displaystyle\lim_{n\to\infty}\tilde x_n’=\tilde x’$. Now,
\begin{align*}
\tilde d(\tilde x,\tilde x’)&=\lim_{n\to\infty}\tilde d(\tilde x_n,\tilde x_n’)\\
&=\lim_{n\to\infty}\hat d(\hat x_n,\hat x_n’)\ (\rho\ \mbox{is an isometry})\\
&=\hat d(\hat x,\hat x)\\
&=0.
\end{align*}
Hence, $\tilde x=\tilde x’$.

Step B. $\psi$ is onto.

Let $\tilde x\in\tilde X$. Then $\exists$ a sequence $(\tilde x_n)$ in $\tilde W$ such that $\displaystyle\lim_{n\to\infty}\tilde x_n=\tilde x$. $(\tilde x_n)$ is Cauchy (since it is a convergent sequence) and $\rho^{-1}$ is an isometry, so the sequence $(\hat x_n)\subset \hat X$, where $\hat x_n=\rho^{-1}\tilde x_n$, is Cauchy. Since $\hat X$ is complete, $\exists\hat x\in\hat X$ such that $\displaystyle\lim_{n\to\infty}\hat x_n=\hat x$. Clearly $\psi\hat x=\tilde x$ and hence $\psi$ is onto.

Step C. $\psi$ is an isometry.

Let $\hat x,\hat y\in\hat X$. Then $\exists$ sequences $(\hat x_n)$, $(\hat y_n)$ in $W$ such that $\displaystyle\lim_{n\to\infty}\hat x_n=\hat x$ and $\displaystyle\lim_{n\to\infty}\hat y_n=\hat y$, respectively.
\begin{align*}
\hat d(\hat x,\hat y)&=\lim_{n\to\infty}\hat d(\hat x_n,\hat y_n)\\
&=\lim_{n\to\infty}\tilde d(\tilde x_n,\tilde y_n)\ (\tilde x_n:=\rho\hat x_n,\ \tilde y_n:=\rho y_n)\\
&=\tilde d(\tilde x,\tilde y)\ (\lim_{n\to\infty}\tilde x_n=\tilde x,\ \lim_{n\to\infty}\tilde y_n=\tilde y)\\
&=\tilde d(\psi\hat x,\psi\hat y).
\end{align*}
Thus, $\psi$ is an isometry.

Remember that an isometry from a metric space into another metric space is automatically one-to-one. Therefore, $\hat X$ is isometric with $\tilde X$ via $\psi$.

Intuitively speaking, the completion of a metric space $X$ can be achieved by adding to $X$ all its limit points. Recall that if $x$ is a limit point of $X$, then there exists a sequence $(x_n)$ in $X$ such that $\displaystyle\lim_{n\to\infty}x_n=x$. This is a reminiscence of extending from rational numbers $\mathbb{Q}$ to real numbers $\mathbb{R}$ (which is complete) by adding to $\mathbb{Q}$ all its limit points (irrational numbers).

# Group Theory 6: Lagrange’s Theorem

In this lecture, we study Lagrange’s Theorem which is named after the Italian/French mathematician, physicists and astronomer Joseph-Louis Lagrange. It states that the order of a subgroup $H$ of a finite group $G$ divides the order of $G$. Lagrange’s Theorem has many important applications in group theory. Before we discuss the theorem we first need to study an important class of binary relations called equivalence relations.

Definition. A binary relation $R$ on a nonempty set $S$ is a subset $R\subset S\times S$. If $(a,b)\in R$, we say $a$ is $R$-related to $b$ and write $aRb$. A binary relation $R$ on a set $S$ is called an equivalence relation on $S$ if it satisfies:

1. $aRa$ $\forall a\in S$. ($R$ is reflexive.)

2. $\forall a,b,\in S$, $aRb\Longrightarrow bRa$. ($R$ is symmetric.)

3. $\forall a,b,c\in S$, $aRb$ and $bRc$ $\Longrightarrow$ $aRc$. ($R$ is transitive.)

Examples. 1. Define a binary relation $\equiv\mod n$ on $\mathbb{Z}$, the set of integers by
$$\forall a,b\in\mathbb{Z},\ a\equiv b\mod n\ \mbox{if}\ n|(a-b).$$ Then $\equiv\mod n$ is an equivalence relation in $\mathbb{Z}$, called the congruence relation modulo $n$. If $a\equiv b\mod n$, we say $a$ is congruent to $b$ modulo $n$. Note that $n|(a-b)$ if and only if $a-b=nk$ for some $k\in\mathbb{Z}$ if and only if $a$ and $b$ have the same remainder when they are divided by $n$ if and only if $a-b\in n\mathbb{Z}$ where
$$n\mathbb{Z}=\{nk: k\in\mathbb{Z}\},$$ the set of all integer multiples of $n$. $\forall n\in\mathbb{Z}$, $n\mathbb{Z}$ is a subgroup of the additive group $(\mathbb{Z},+)$.

2. Let $G$ be a group and $H\leq G$. Define a binary relation $R$ on $G$ by
$$\forall a,b\in G,\ aRb\ \mbox{if}\ ab^{-1}\in H.$$ Then $R$ is an equivalence relation on $G$.

Definition. If $R$ is an equivalence relation on a set $S$, then $[a]$, the equivalence class of $a\in S$, is defined by
$$[a]=\{b\in S: bRa\}.$$

Theorem. Let $R$ be an equivalence relation on a set $S$. Then

1. $S=\bigcup_{a\in S}[a]$.

2. $\forall a,b\in S$, either $[a]=[b]$ or $[a]\cap [b]=\emptyset$.

This means that an equivalence relation $R$ on a set $S$ partitions $S$ into equivalence classes. The set of all equivalence classes $\{[a]: a\in S\}$ is denoted by $S/R$ and is called the quotient set of $S$ modulo $R$.

The converse of the above theorem is also true, namely

Theorem. Let $\wp=\{A_i:i\in I\}$ be a partition of a set $S$ i.e. $S=\bigcup_{i\in I}A_i$ and $\forall i\ne j\in I$, $A_i\cap A_j=\emptyset$. Define a binary relation $R$ on $S$ by
$$\forall a,b\in S,\ aRb\ \mbox{if}\ a,b\in A_i\ \mbox{for some}\ i\in I.$$ Then $R$ is an equivalence relation on $S$, called the equivalence relation induced by the partition $\wp$.

Example. Let $f: X\longrightarrow Y$ be a surjective map. Then $\{f^{-1}(y): y\in Y\}$ forms a partition of $X$. The equivalence relation induced by this partition is
$$\ker f=\{(a,b)\in X\times X: f(a)=f(b)\}.$$ This equivalence relation is called the kernel of the function $f$.

Let $G$ be a group and $H\leq G$. Consider the equivalence relation $R$ on $G$ given by
$$\forall a,b\in G,\ aRb\ \mbox{if}\ ab^{-1}\in H.$$
Then for each $a\in G$, $[a]=Ha$ where $Ha=\{ha:h\in H\}$. $Ha$ is called a right coset of $H$ in $G$.

Example. Consider $\equiv\mod n$, the congruence relation modulo $n$ on $\mathbb{Z}$. For each $a\in \mathbb{Z}$, $[a]=n\mathbb{Z}+a$. The right coset $n\mathbb{Z}+a$ is called the congruence class of $a$ modulo $n$.

Now we are ready to discuss Lagrange’s Theorem.

Theorem [Lagrange's Theorem] If $G$ is a finite group and $H\leq G$, then $|H|||G|$.

Proof. $G=Ha_1\cup Ha_2\cup\cdots\cup Ha_k$ and $Ha_i\cup Ha_j=\emptyset$ if $i\ne j$, $i,j=1,2,\dots, k$. Due to the cancellation law, we see that $|H|=|Ha_i|$ for all $i=1,2,\cdots, k$. Therefore,
\begin{align*}
|G|&=\sum_{i=1}^k|Ha_i|\\
&=\sum_{i=1}^k|H|\\
&=k|H|.
\end{align*}
This completes the proof.

If $G$ is finite, the number of right cosets of $H$ in $G$, namely $\frac{|G|}{|H|}$ is called the index of $H$ in $G$ and is written as $|G:H|$ or $i_G(H)$. Here, we use the notation $|G:H|$.

It should be noted that the converse of Lagrange’s Theorem need not be true. For example, the alternating group of index 4
$$A_4=\{1,(123),(132),(124),(142),(134),(143),(234),(243),(12)(34),(13)(24),(14)(23)\}$$ has no subgroup of order 6 though $|A_4|=12$ and $6|12$.

Theorem. A group $G$ of prime order is cyclic.

Proof. Let $|G|=p$, a prime. Let $H\leq G$. Then by Lagrange’s Theorem, $|H|||G|$ and so either $|H|=1$ or $|H|=p$. If $H\ne\{e\}$, then $H=G$. If $a\ne e\in G$, $\langle a\rangle=G$.

Let $G$ be a finite group and $a\ne e\in G$. Then $H=\langle a\rangle\leq G$. So, $a^m=e$ for some $m\in\mathbb{N}$.

Definition. If $G$ is finite, then the order of $a\ne e\in G$, written $|a|$ or $o(a)$, is the least positive integer $n$ such that $a^n=e$.

Corollary. If $G$ is finite and $a\ne e\in G$, then $|\langle a\rangle|||G|$.

Theorem. If $G$ is a finite group of order $n$, then $a^n=e$ for all $a\in G$.

Proof. Let $a\ne e\in G$ and $|\langle a\rangle|=k$. Then by Lagrange’s Theorem $k|n$ i.e. $n=kl$ for some $l\in\mathbb{N}$. So,
$$a^n=a^{kl}=(a^k)^l=e^l=e.$$

# Group Theory 5: Subgroups

In this lecture, we study subgroups. A nonempty subset $H$ of a group $G$ is called a subgroup of $G$ if $H$ itself forms a group relative to the operation $\cdot$ in $G$. If $H$ is a subgroup of $G$, we simply write $H\leq G$. Clearly $G\leq G$ and $\{e\}\leq G$. $G$ and $\{e\}$ are called trivial subgroups of $G$. Before we discuss subgroups further we need to go over some basic properties of a group as we need them.

Lemma. If $G$ is a group, then:

(a) Its identity element is unique.

(b) Each $a\in G$ has a unique inverse $a^{-1}\in G$.

(c) $\forall a\in G$, $(a^{-1})^{-1}=a$.

(d) $\forall a,b\in G$, $(ab)^{-1}=b^{-1}a^{-1}$.

The proof of this lemma is pretty much straightforward and is left to readers.

In order to show that a nonempty subset $H$ of a group $G$ is a subgroup we need to check:

1. $H$ is closed under $\cdot$, the operation in $G$. That is, $\forall a,b\in H$, $ab\in H$.

2. $e\in H$.

3. $\forall a\in H$, $a^{-1}\in H$.

Note that the associative law holds automatically. It turns out that there is a simpler criterion to check if a nonempty subset is a subgroup of a group.

Lemma. A nonempty subset $H\subset G$ is a subgroup of $G$ if and only if $\forall a,b\in H$, $ab^{-1}\in H$.

Proof. ($\Longrightarrow$) Clear.

($\Longleftarrow$) Let $a,b\in H$. Then by assumption, $aa^{-1}=e\in H$. Since $a,e\in H$, $ea^{-1}=a^{-1}\in H$. Since $b^{-1}\in H$, $a(b^{-1})^{-1}=ab\in H$. Therefore, $H$ is a subgroup of $G$.

Example. $\forall n\in\mathbb{N}\cup\{0\}$, $n\mathbb{Z}\leq(\mathbb{Z},+)$. Here, $n\mathbb{Z}=\{nx: x\in\mathbb{Z}\}$. In fact, they are all the subgroups of $(\mathbb{Z},+)$.

Definition. The cyclic subgroup of $G$ generated by $a\in G$ is $\{a^k: k\in\mathbb{Z}\}$. It is denoted by $\langle a\rangle$.

Example. Let $G$ be a group and $a\in G$. Let $C(a)=\{g\in G: ga=ag\}$. Then $C(a)\leq G$. $C(a)$ is called the centralizer of $a$ in $G$.

Example. Let $G$ be a group and let $Z(G)=\{z\in G: zx=xz\ \forall x\in G\}$. Then $Z(G)$ is a subgroup of $G$, called the center of $G$.

Example. Let $G$ be a group and $H\leq G$. $\forall a\in G$, $a^{-1}Ha=\{a^{-1}ha: h\in H\}$ is also a subgroup of $G$.

Lemma. Suppose that $(G,\cdot)$ is a group and $H$ is a finite nonempty subset of $H$. If $H$ is closed under $\cdot$, then $H$ is a subgroup of $G$.

Proof. Let $H=\{a_1,a_2,\cdots,a_n\}$. Then $\forall a\in H$, $aH=\{aa_1,aa_2,\cdots,aa_n\}\subset H$. On the other hand, if $aa_i=aa_j$ then $a_i=a_j$. So, $|aH|=|H|$. This implies that $aH=H$ and hence
\begin{align*}
a\in aH &\Longrightarrow a=aa_i\ \mbox{for some}\ i\\
&\Longrightarrow a_i=e\in H.
\end{align*}
Since $e\in aH$, $e=aa_j$ for some $j$ $\Longrightarrow$ $a_j=a^{-1}$.

Example. The symmetric group $S_3=\{1,(123),(132),(23),(13),(12)\}$ has 5 subgroups $\{1\}$, $\{1,(12)\}$, $\{1,(13)\}$, $\{1,(23)\}$, and $A_3=\{1,(123),(132)\}$. $A_3$ is called alternating group of degree 3. Cycles of length 2 such as (12), (13), (23) are called transpositions. Any permutation can be written as a product of either an even number of transpositions or an odd number of transpositions. If a permutation is a product of an even number of transpositions, the permutation is called an even permutation. If a permutation is a product of an odd number of transpositions, the permutation is called an odd permutation. The identity permutation 1 is an even permutation as $\begin{pmatrix} 1 & 2 & 3\\ 1 & 2 & 3 \end{pmatrix}$ can be written as (12)(21)(23)(32)(31)(13). The permutation (123) can be written as (12)(13). So, (123) is an even permutation. Similarly (132) is an even permutation as well. So it turns out that the alternating group $A_3$ is the set of all even permutations. In general, the set $A_n$ of all even permutations of $\{1,2,3,\cdots,n\}$ form a subgroup of the symmetric group $S_n$. Remember that $S_n$ has order $n!$. Since there are exactly the same number of even permutations and odd permutations, the alternating group $A_n$ of degree $n$ has order $\frac{n!}{2}$.

Example. $\mathrm{SL}(n,\mathbb{R})=\{A\in\mathrm{GL}(n,\mathbb{R}): \det A=1\}$ is a subgroup of the general linear group $\mathrm{GL}(n,\mathbb{R})$ of degree $n$. $\mathrm{SL}(n,\mathbb{R})$ is called the special linear group of degree $n$. The general linear group $\mathrm{GL}(n,\mathbb{R})$ is the group of isometries of $\mathbb{R}^n$. An isometry of $\mathbb{R}^n$ is a linear bijective map from $\mathbb{R}^n$ onto itself which preserves the inner product. What this means is that if $T: \mathbb{R}^n\longrightarrow\mathbb{R}^n$ is an isometry then for any two vectors $v,w\in\mathbb{R}^n$, $\langle Tv,Tw\rangle=\langle v,w\rangle$. The special linear group $\mathrm{SL}(n,\mathbb{R})$ is the group of all orientation preserving isometries of $\mathbb{R}^n$.

# Functional Analysis 4: Convergence, Cauchy Sequence, Completeness

The set $\mathbb{Q}$ of rational numbers is not complete (or not a continuum) since it has gaps or holes. For instance, $\sqrt{2}$ is not in $\mathbb{Q}$. On the other hand, the set $\mathbb{R}$ of real numbers has no gaps or holes, so it is complete (or is a continuum). Let $(x_n)$ be a sequence of real numbers. Suppose that $(x_n)$ converges to a real number $x$. Then by the triangle inequality, for any $m,n\in\mathbb{N}$, we have
$$|x_m-x_n|\leq |x_m-x|+|x-x_n|.$$
Hence, $\displaystyle\lim_{m,n\to\infty}|x_m-x_n|=0$, i.e. $(x_n)$ is a Cauchy sequence. Conversely, Georg Cantor introduced the completeness axiom that every Cauchy sequence of real numbers converges and defined a real number as the limit of a Cauchy sequence of rational numbers. For instance, consider the Cauchy sequence $(x_n)$ defined by
$$x_1=1,\ x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n},\ \forall n\geq 2.$$
If $(x_n)$ converges to a number $x$, it would satisfy $x^2=2$ i.e. $(x_n)$ converges to $\sqrt{2}$. There is another way to obtain the completeness of $\mathbb{R}$ by a Dedekind cut, though we are not going to delve into that here.

More generally, one can also consider a complete metric space and that is what we are going to study in this lecture.

Definition. A sequence $(x_n)$ is a metric space $(X,d)$ is said to converge or to be convergent to $x\in X$ if
$$\lim_{n\to\infty}d(x_n,x)=0.$$
$x$ is called the limit if $(x_n)$ and we write
$$\lim_{n\to\infty}x_n=x\ \mbox{or}\ x_n\rightarrow x.$$
If $(x_n)$ is not convergent, it is sad to be divergent. We can generalize the definiton of the convergence of a sequence we learned in calculus in terms of a metric as:

Definition. $\displaystyle\lim_{n\to\infty}d(x_n,x)=0$ if and only if given $\epsilon>0$ $\exists$ a positive integer $N$ s.t. $x_n\in B(x,\epsilon)$ $\forall n\geq N$.

A nonempty subset $M\subset X$ is said to be bounded if
$$\delta(M)=\sup_{x,y\in M}d(x,y)<\infty.$$

Lemma. Let $(X,d)$ be a metric space.

(a) A convergent sequence in $X$ is bounded and its limit is unique.

(b) If $x_n\rightarrow x$ and $y_n\rightarrow y$, then $d(x_n,y_n)\rightarrow d(x,y)$.

Proof. (a) Suppose that $x_n\rightarrow x$. Then one can find a positive integer $N$ such that $d(x_n,x)<1$ $\forall n\geq N$. Let $M=2\max\{d(x_1,x),\cdots,d(x_{N-1},x),1\}$. Then for all $m,n\in\mathbb{N}$,
\begin{align*}
d(x_m,x_n)&\leq d(x_m,x)+d(x,x_n)\ (\mbox{ (M3) triangle inequality)}\\
&\leq M.
\end{align*}
This means that $\delta((x_n))\leq M<\infty$ i.e. $(x_n)$ is bounded.

Suppose that $x_n\rightarrow x$ and $x_n\rightarrow y$. Then
\begin{align*}
0\leq d(x,y)&\leq d(x,x_n)+d(x_ny)\\
&\rightarrow 0
\end{align*}
as $n\to\infty$. So, $d(x,y)=0\Rightarrow x=y$ by (M1).

(b) By (M3),
$$d(x_n,y_n)\leq d(x_n,x)+d(x,y)+d(y,y_n)$$
and so we obtain
$$d(x_n,y_n)-d(x,y)\leq d(x_n)+d(y,y_n).$$
Similarly, we also obtain the inequality
$$d(x,y)-d(x_n,y_n)\leq d(x,x_n)+d(y_n,y).$$
Hence,
$$0\leq |d(x_n,y_n)-d(x,y)|\leq d(x_n,x)+d(y_n,y)\rightarrow 0$$
as $n\to\infty$.

Definition. A sequence $(x_n)\subset (X,d)$ is said to be Cauchy if given $\epsilon>0$ $\exists$ a positive integer $N$ such that
$$d(x_m,x_n)<\epsilon\ \forall m,n\geq N.$$
The space $X$ is said to be complete if every Cauchy sequence in $X$ converges.

Examples. The real line $\mathbb{R}$ and the complex plane $\mathbb{C}$ are complete.

Theorem. Every convergent sequence is Cauchy.

Proof. Suppose that $x_n\rightarrow x$. Then given $\epsilon>0$ $\exists$ a poksitive integer $N$ s.t. $d(x_n,x)<\frac{\epsilon}{2}$ for all $n\geq N$. Now, $\forall m,n\geq N$
$$d(x_m,x_n)\leq d(x_m,x)+d(x,x_n)<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$$
Therefore, $(x_n)$ is Cauchy.

Theorem. Let $M$ be a nonempty subset of a metric space $(X,d)$. Then

(a) $x\in\bar M\Longleftrightarrow \exists$ a seqence $(x_n)\subset M$ such that $x_n\rightarrow x$.

(b) $M$ is closed $\Longleftrightarrow$ given a sequence $(x_n)\subset M$, $x_n\rightarrow x$ implies $x\in M$.

Proof. (a) ($\Longrightarrow$) Since $x\in\bar M$, $\forall n\in\mathbb{N}$ $\exists x_n\in B\left(x,\frac{1}{n}\right)\cap M\ne\emptyset$. Let $\epsilon>0$ be given. Then by the Archimedean property, $\exists$ a positive integer $N$ s.t. $N\geq\frac{1}{\epsilon}$. Now,
$$n\geq N\Longrightarrow d(x_n,x)<\frac{1}{n}\leq\frac{1}{N}<\epsilon.$$

($\Longleftarrow$) Suppose that $\exists$ a sequence $(x_n)\subset M$ s.t. $x_n\rightarrow x$. Then given $\epsilon>0$ $\exists$ a positive integer $N$ s.t. $x_n\in B(x,\epsilon)$ $\forall n\geq N$. This means that $\forall\epsilon>0$, $B(x,\epsilon)\cap M\ne\emptyset$. So, $x\in\bar M$.

(b) ($\Longrightarrow$) Clear

($\Longleftarrow$) It suffices to show that $\bar M\subset M$. Let $x\in\bar M$. Then $\exists$ a sequence $(x_n)\subset M$ such that $x_n\rightarrow x$. By assumption, $x\in M$.
Theorem. A subspace $M$ of a complete metric space $X$ itself is complete if and only if $M$ is closed in $X$.

Proof. ($\Longrightarrow$) Let $M\subset X$ be complete. Let $(x_n)$ be a sequence in $M$ such that $x_n\rightarrow x$. Then $(x_n)$ is Cauchy. Since $M$ is complete, every Cauchy sequence must converge and hence $x\in M$. This means that $M$ is closed.

($\Longleftarrow$) Suppose that $M\subset X$ is closed. Let $(x_n)$ be a Cauchy sequence in $M\subset X$. Since $X$ is complete, $\exists x\in X$ such that $x_n\rightarrow x$. Since $M$ is closed, $x\in M$. Therefore, $M$ is complete.

Example. In $\mathbb{R}$ with Euclidean metric, the closed intervals $[a,b]$ are complete. $\mathbb{Z}$, the set of integers is also complete by the above theorem since it is closed in $\mathbb{R}$. One can directly see why $\mathbb{Z}$ is complete without quoting the theorem though. Let $(x_n)$ be a Cauchy sequence in $\mathbb{Z}$. Then we see that there exists a positive integer $N$ such that $x_N=x_{N+1}=x_{N+2}=\cdots$. Hence any Cauchy sequence in $\mathbb{Z}$ is a convergent sequence in $\mathbb{Z}$. Therefore, $\mathbb{Z}$ is complete.

Theorem. A mapping $T: X\longrightarrow Y$ is continuous at $x_0\in X$ if and only if $x_n\rightarrow x$ implies $Tx_n\rightarrow Tx_0$.

Proof. ($\Longrightarrow$) Suppose that $T$ is continuous and $x_n\rightarrow x$ in $X$. Let $\epsilon>0$ be given. Then $\exists\delta>0$ s.t. whenever $d(x,x_0)<\delta$, $d(Tx,Tx_0)<\epsilon$. Since $x_n\rightarrow x$, $\exists$ a positive integer $N$ s.t. $d(x_n,x_0)<\delta$ $\forall n\geq N$. So, $\forall n\geq N$, $d(Tx_n,Tx_0)<\epsilon$. Hence, $Tx_n\rightarrow Tx_0$.

($\Longleftarrow$) Suppose that $T$ is not continuous. Then $\exists\epsilon>0$ s.t. $\forall\delta>0$, $\exists x\ne x_0$ satisfying $d(x,x_0)<\delta$ but $d(Tx,tx_0)\geq\epsilon$. So, $\forall n=1,2,\cdots$, $\exists x_n\ne x_0$ satisfying $d(x_n,x_0)<\frac{1}{n}$ but $d(Tx_n,Tx_0)\geq\epsilon$. This means that $x_n\rightarrow x_0$ but $Tx_n\not\rightarrow Tx_0$.