# Functions of a Complex Variable 6: Harmonic Functions

Throughout this course, a connected open subset of $\mathbb{C}$ is called a domain. Suppose that a function $f(z)=u(x,y)+iv(x,y)$ is analytic in a domain $\mathcal{D}$. Then $f(z)$ satisfies the Cauchy-Riemann equations i.e.

\label{eq:c-r}
u_x=v_y,\ u_y=-v_x.

Differentiating the Cauchy-Riemann equations \eqref{eq:c-r} with respect to $x$, we obtain

\label{eq:c-r2}
u_{xx}=v_{yx},\ u_{yx}=-v_{xx}.

Differentiating the Cauchy-Riemann equations \eqref{eq:c-r} with respect to $y$, we obtain

\label{eq:c-r3}
u_{xy}=v_{yy},\ u_{yy}=-v_{xy}.

By the continuity of the partial derivatives of $u(x,y)$ and $v(x,y)$, we have

\label{eq:cont}
u_{xy}=u_{yx},\ v_{xy}=v_{yx}.

Applying \eqref{eq:cont} to \eqref{eq:c-r2} and \eqref{eq:c-r3}, we obtain the Laplace equations:
$$u_{xx}+u_{yy}=0,\ v_{xx}+v_{yy}=0.$$
That is to say, $u(x,y)$ and $v(x,y)$ are harmonic maps in $\mathcal{D}$.

Example. The function $f(z)=e^{-y}\sin x-ie^{-y}\cos x$ is entire (i.e analytic on the complex plane $\mathbb{C}$), so both $e^{-y}\sin x$ and $-e^{-y}\cos x$ are harmonic. (You can of course check it for yourself!)

If two functions $u(x,y)$, $v(x,y)$ are harmonic in a domain $\mathcal{D}$ and their first-order partial derivatives satisfy the Cauchy-Riemann equations \eqref{eq:c-r} throughout $\mathcal{D}$, $v(x,y)$ is said to be a harmonic conjugate of $u(x,y)$.

Theorem. A function $f(z)=u(x,y)+iv(x,y)$ is analytic in a domain $\mathcal{D}$ if and only if $v(x,y)$ is a harmonic conjugate of $u(x,y)$.

Remark. If $v(x,y)$ is a harmonic conjugate of $u(x,y)$ in some domain, it is not in general true that $u$ is a harmonic conjugate of $v$ there.

Example. Let $u(x,y)=x^2-y^2$ and $v(x,y)=2xy$. Since $f(z)=z^2=(x^2-y^2)+i2xy$ is entire, $v(x,y)$ is a harmonic conjugate of $u(x,y)$. However, $u(x,y)$ cannot be a harmonic conjugate of $v(x,y)$ since $2xy+i(x^2-y^2)$ is not analytic anywhere.

Example. [Finding a harmonic conjugate of a harmonic function] Let $u(x,y)=y^3-3x^2y$ and $v(x,y)$ be a harmonic conjugate of $u(x,y)$. Then it follows from the Cauchy-Riemann equations \eqref{eq:c-r} that $v_y=-6xy$. Integrating this with respect to $y$, we obtain
$$v(x,y)=-3xy^2+\phi(x),$$
where $\phi(x)$ is some unknown function. We determine $\phi(x)$ using $u_y=-v_x$:
$$v_x=-3y^2+\phi’(x).$$
Comparing this with $-u_y=-3y^2+3x^2$, we get $\phi’(x)=3x^2$ and so, $\phi(x)=x^3+C$ where $C$ is a constant. Hence, we find a harmonic conjugate of $u(x,y)$:
$$v(x,y)=-3xy^2+x^3+C,$$
where $C$ is a constant. The corresponding analytic function is
$$f(z)=(y^3-3x^2y)+i(-3xy^2+x^3+C),$$
where $C$ is a constant.

# Group Theory 13: Finitely Generated Abelian Groups

The group $\mathbb{Z}\times\mathbb{Z}_2$ is generated by $(1,0)$ and $(0,1)$. In general, the direct product of $n$ cyclic groups, each of which is either $\mathbb{Z}$ or $\mathbb{Z}_m$ is generated by $(1,0,\cdots,0)$, $(0,1,0,\cdots,0)$, $\cdots$, $(0,0,\cdots,0,1)$. Such a direct product may be generated by fewer elements. For example, $\mathbb{Z}_3\times\mathbb{Z}_4\times\mathbb{Z}_{35}$ is generated by the single element $(1,1,1)$ i.e. it is a cyclic group. Conversely, we have the following theorem holds.

Theorem. [Fundamental Theorem of Finitely Generated Abelian Groups] Every finitely generated abelian group $G$ is isomorphic to a direct product of cyclic groups in the form
$$\mathbb{Z}_{p_1^{r_1}}\times\mathbb{Z}_{p_2^{r_2}}\times\cdots\times\mathbb{Z}_{p_n^{r_n}}\times\mathbb{Z}\times\mathbb{Z}\times\cdots\times\mathbb{Z},$$where the $p_i$ are primes, not necessarily distinct and the $r_i$ are positive integers. The direct product is unique except for a possible rearrangement of the factors. The number of factors $\mathbb{Z}$ is called the Betti number of $G$.

Example. Find all abelian groups, up to isomorphism, of order 360.

Solution. $360=2^3\cdot 3^3\cdot 5$, so all abelian groups, up to isomorphism, of order 360 are
\begin{align*}
\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_3\times\mathbb{Z}_3\times\mathbb{Z}_5,\\
\mathbb{Z}_2\times\mathbb{Z}_4\times\mathbb{Z}_3\times\mathbb{Z}_3\times\mathbb{Z}_5,\\
\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_9\times\mathbb{Z}_5,\\
\mathbb{Z}_2\times\mathbb{Z}_4\times\mathbb{Z}_9\times\mathbb{Z}_5,\\
\mathbb{Z}_8\times\mathbb{Z}_3\times\mathbb{Z}_3\times\times\mathbb{Z}_5,\\
\mathbb{Z}_8\times\mathbb{Z}_9\times\mathbb{Z}_5.
\end{align*}

Definition. A group $G$ is decomposable if it is isomorphic to a direct product of two proper nontrivial subgroups. Otherwise $G$ is indecomposable.

Theorem. The finite indecomposable abelian groups are exactly the cyclic groups with order a power of a prime.

Proof. If $G$ is a finite indecomposable abelian group, then by Fundamental Theorem of Finitely Generated Abelian Groups, $G$ is isomorphic to a direct product of cyclic groups of prime power order.Since $G$ is indecomposable, the direct product must consist of just one cyclic group of a prime power order. conversely, let $p$ be a prime. Then $\mathbb{Z}_{p^r}$ is indecomposable. If it were isomorphic to $\mathbb{Z}_{p^i}\times\mathbb{Z}_{p^j}$ where $i+j=r$, then every element has an order at most $p^{\max(i,j)}<p^r$.

Theorem. If $m$ divides the order of a finite abelian group $G$, then $G$ has a subgroup of order $m$.

Proof. Since $G$ is a finite abelian group,
$$G\cong\mathbb{Z}_{p_1^{r_1}}\times\mathbb{Z}_{p_2^{r_2}}\times\cdots\times\mathbb{Z}_{p_n^{r_n}}.$$Since $p_1^{r_1}p_2^{r_2}\cdots p_n^{r_n}=|G|$, $m$ must be of the form $p_1^{s_1}p_2^{s_2}\cdots p_n^{s_n}$, where $0\leq s_i\leq r_i$. For each $1\leq i\leq n$, $p_i^{r_i-s_i}$ generates a cyclic group of order$$\frac{p_i^{r_i}}{(p_i^{r_i},p_i^{r_i-s_i})}=\frac{p_i^{r_i}}{p_i^{r_i-s_i}}=p_i^{s_i}.$$So, $p_i^{r_i-s_i}$ generates a cyclic subgroup of $\mathbb{Z}_{p_i^{r_i}}$ of order $p_i^{s_i}$. Therefore,$$\langle p_1^{r_1-s_1}\rangle\times\langle p_2^{r_2-s_2}\rangle\times\cdots\times\langle p_n^{r_n-s_n}\rangle$$
is a subgroup of $G$ of order $m=p_1^{s_1}p_2^{s_2}\cdots p_n^{s_n}$.

Theorem. If $m$ is a square free integer i.e. if $m$ is not divisible by the square of any prime, then every abelian group of order $m$ is cyclic.

Proof. Let $G$ be an abelian group of square free order $m$. Then
$$G\cong\mathbb{Z}_{p_1^{r_1}}\times\mathbb{Z}_{p_2^{r_2}}\times\cdots\times\mathbb{Z}_{p_n^{r_n}},$$
where $m=p_1^{r_1}p_2^{r_2}\cdots p_n^{r_n}$. Since $m$ is square free, $r_i=1$, $i=1,\cdots,n$ and the $p_i$ are distinct. Hence, $G$ is isomorphic to $\mathbb{Z}_{p_1p_2\cdots p_n}$ i.e. it is cyclic.

# Group Theory 12: Direct Products

Let $G_1,G_2,\cdots,G_n$ be groups. Consider the Cartesian product
$$\prod_{i=1}^nG_i:=G_1\times G_2\times\cdots\times G_n.$$
Define a binary operation $\cdot$ on $\prod_{i=1}^nG_i$ by
$$(a_1,a_2,\cdots,a_n)\cdot(b_1,b_2,\cdots,b_n)=(a_1b_1,a_2b_2,\cdots,a_nb_n)$$
for $(a_1,a_2,\cdots,a_n),(b_1,b_2,\cdots,b_n)\in\prod_{i=1}^nG_i$. Then $\cdot$ is well-defined. Clearly $\cdot$ is associative. $(e_1,e_2,\cdots,e_n)\in\prod_{i=1}^nG_i$ is an identity element. For each $(a_1,a_2,\cdots,a_n)\in\prod_{i=1}^nG_i$, $(a_1,a_2,\cdots,a_n)^{-1}=(a_1^{-1},a_2^{-1},\cdots,a_n^{-1})\in\prod_{i=1}^nG_i$. So, $\left(\prod_{i=1}^nG_i,\cdot\right)$ is a group called the direct product of the $G_i$’s. If the operation on each $G_i$ is commutative, refer to $\prod_{i=1}^nG_i$ as the direct sum of the groups $G_i$. In this case, we often use the notation $\bigoplus_{i=1}^nG_i$ instead of $\prod_{i=1}^nG_i$.

Example. Let $\mathbb{R}\oplus\mathbb{R}$ be the direct sum of $(\mathbb{R},+)$ and itself. Define a map $\varphi:\mathbb{R}\oplus\mathbb{R}\longrightarrow S^1\times S^1$ by
$$\varphi(x,y)=(e^{2\pi ix},e^{2\pi iy})$$
for each $(x,y)\in\mathbb{R}\oplus\mathbb{R}$. Then $\varphi$ is an onto-homomorphism. The kernel of $\varphi$ is
$$\ker\varphi=\mathbb{Z}\oplus\mathbb{Z}.$$
Hence, by the Fundamental Homomorphism Theorem
$$\mathbb{R}\oplus\mathbb{R}/\mathbb{Z}\oplus\mathbb{Z}\cong S^1\times S^1.$$That is, the quotient group $\mathbb{R}\oplus\mathbb{R}/\mathbb{Z}\oplus\mathbb{Z}$ is a torus. $\mathbb{R}\oplus\mathbb{R}/\mathbb{Z}\oplus\mathbb{Z}$ can be viwed as a quotient set $\mathbb{R}\oplus\mathbb{R}/\sim$ where $\sim$ is an equivalence relation on $\mathbb{R}\oplus\mathbb{R}$ defined as follows: For all $(x,y),(z,w)\in\mathbb{R}\oplus\mathbb{R}$,$$(x,y)\sim (z,w)\ \mbox{if}\ (x,y)-(z,w)=(m,n)$$for some $(m,n)\in\mathbb{Z}\times\mathbb{Z}$.

The following theorem is introduced without a proof.

Theorem. Let $(a_1,\cdots,a_n)\in\prod_{i=1}^nG_i$. If for each $i=1,\cdots,n$, $a_i$ is of finite order $r_i$ in $G_i$, then the order of $(a_1,\cdots,a_n)$ in $\prod_{i=1}^nG_i$ is the least common multiple of $r_1,r_2,\cdots,r_n$.

Example. Find the order of $(8,4,10)$ in $\mathbb{Z}_{12}\times\mathbb{Z}_{60}\times\mathbb{Z}_{24}$.

Solution. First we find the orders of 8, 4, 10 in $\mathbb{Z}_{12}$, $\mathbb{Z}_{60}$, and $\mathbb{Z}_{24}$, respectively. For that let us recall a theorem we studied here. The theorem can be restated for an additive group as:

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

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

Since 1 has order $n$ in $\mathbb{Z}_n$, we have the following corollary.

Corollary. The order of $1\leq k\leq n-1$ in $\mathbb{Z}_n$ is $\frac{n}{(k,n)}$.

It follows from this corollary that the order of 8 in $\mathbb{Z}_{12}$ is $\frac{12}{(8,12)}=\frac{12}{4}=3$, the order of 4 in $\mathbb{Z}_{60}$ is $\frac{60}{(4,60)}=\frac{60}{4}=15$, and the order of 10 in $\mathbb{Z}_{24}$ is $\frac{24}{(10,24)}=\frac{24}{2}=12$. The least common multiple of 3, 15, 12 is 60, so the order of $(8,4,10)$ in $\mathbb{Z}_{12}\times\mathbb{Z}_{60}\times\mathbb{Z}_{24}$ is 60.

Example. $\mathbb{Z}_2\times\mathbb{Z}_3=\{(0,0), (0,1), (0,2),(1,0),(1,1),(1,3)\}$ is a cyclic group generated by $(1,1)$. Hence, $\mathbb{Z}_2\times\mathbb{Z}_3\cong\mathbb{Z}_6$.

Example. $\mathbb{Z}_2\times\mathbb{Z}_2$ is not cyclic. $\mathbb{Z}_2\times\mathbb{Z}_2\cong V_4$, Klein four-group.

Theorem. $\mathbb{Z}_m\times\mathbb{Z}_n$ is cyclic and isomorphic to $\mathbb{Z}_{mn}$ if and only if $(m,n)=1$.

Corollary. $\prod_{i=1}^n\mathbb{Z}_{m_i}$ is cyclic and isomorphic to $\mathbb{Z}_{m_1m_2\cdots m_n}$ if and only if the numbers $m_i$ for $i=1,2,\cdots,n$ are such that the greatest common divisor of any two of them is 1.

Corollary. If $n=(p_1)^{n_1}(p_2)^{n_2}\cdots(p_r)^{n_r}$ where $p_1,p_2,\cdots,p_r$ are distinct primes, then
$$\mathbb{Z}_n\cong\mathbb{Z}_{(p_1)^{n_1}}\times\mathbb{Z}_{(p_2)^{n_2}}\times\cdots\times\mathbb{Z}_{(p_r)^{n_r}}.$$

Example. $\mathbb{Z}_8\times\mathbb{Z}_9\cong\mathbb{Z}_{72}$.

Let $\prod_{i=1}^nG_i$ be the direct product of groups $G_1,\cdots,G_n$. For each $i=1,\cdots,n$, let
$$\bar G_i=\{(e_1,e_2,\cdots,e_{i-1},a_i,e_{i+1},\dots,e_n): a_i\in G_i\}\leq\prod_{i=1}^n G_i.$$
Then for each $i=1,\cdots,n$, $\bar G_i\cong G_i$. The direct product $\prod_{i=1}^n\bar G_i$ of the groups $\bar G_1,\cdots,\bar G_n$ is called an internal direct product while $\prod_{i=1}^nG_i$ is called an external direct product. Clearly the external and internal direct products are isomorphic to each other.

# Group Theory 11: The Isomorphism Theorems

The following theorem is called the Fundamental Homomorphism Theorem or the First Isomorphism Theorem.

Theorem. Let $G$ and $G’$ be groups, and $\varphi:G\longrightarrow G’$ an epimorphism (onto homomorphism). Then $G/K\cong G’$ where $K=\ker\varphi$.

Proof.

Let $\gamma: G\longrightarrow G/K$ be the canonical homomorphism i.e. for any $a\in G$, $\gamma(a)=Ka$. Define $\psi: G/K\longrightarrow G’$ by
$$\psi(Ka)=\varphi(a)$$for each $a\in G$. Then

1. $\psi$ is well-defined:\begin{align*}Ka=Kb&\Rightarrow ab^{-1}\in K\\&\Rightarrow\varphi(ab^{-1})=e’\\&\Rightarrow\psi(Ka)=\varphi(a)=\varphi(b)=\psi(Kb).\end{align*}
2. $\psi$ is a homomorphism:$$\psi(KaKb)=\psi(Kab)=\varphi(ab)=\varphi(a)\varphi(b)=\psi(Ka)\psi(Kb).$$
3. $\psi$ is one-to-one:\begin{align*}\psi(Ka)=\psi(Kb)&\Rightarrow\varphi(a)=\varphi(b)\\&\Rightarrow\varphi(ab^{-1})=e’\\&\Rightarrow ab^{-1}\in K\\&\Rightarrow Ka=Kb.\end{align*}
4. $\psi$ is onto: Let $b\in G’$. Then there exists $a\in G$ such that $\varphi(a)=b$ since $\varphi$ is onto. Now, $Ka\in G/K$ and $\psi(Ka)=\varphi(a)=b$.

Example. Let $S^1$ be the unit circle centered at the origin. Then it can be represented in terms of complex numbers as
$$S^1=\{e^{2x\pi i}:x\in[0,1)\}.$$Define a map $\varphi: (\mathbb{R},+)\longrightarrow(S^1,\cdot)$ by $$\varphi(x)=e^{2\pi ix}$$for each $x\in\mathbb{R}$. Then $\varphi$ is an onto homomorphism. The kernel of $\varphi$ is$$\ker\varphi=\mathbb{Z}.$$Hence, by the Fundamental Homomorphism Theorem
$$\mathbb{R}/\mathbb{Z}\cong S^1.$$Note that $\mathbb{R}/\mathbb{Z}$ can be viewed as the quotient set $\mathbb{R}/\sim$ where $\sim$ is an equivalence relation on $\mathbb{R}$ defined as follows: For all $x,y\in\mathbb{R}$,
$$x\sim y\ \mbox{if}\ x-y=n$$for some $n\in\mathbb{Z}$.

Theorem [Correspondence Theorem]. Let $\varphi: G\longrightarrow G’$ be a homomorphism. Let $K=\ker\varphi$, $H’\leq G’$ and $H=\varphi^{-1}(H’)=\{a\in G: \varphi(a)\in H\}$. Then $K\subset H\leq G$ and $H/K\cong H’$. If $H’\triangleleft G’$, then $H\triangleleft G’$.

Proof. Since $e\in K\subset H$, $H\ne\emptyset$. Let $a,b\in H$. Then $\varphi(a),\varphi(b)\in H’$. Since $H’\leq G’$, $\varphi(a)\varphi(b)^{-1}=\varphi(ab^{-1})\in H’$ and so, $ab^{-1}\in H$. Hence, $H\leq G$. Since $e’\in H’$, $K\subset H$. Let $\psi=\varphi|_H$. Then $\psi$ is a homomorphism from $H$ onto $H’$ and $\ker\psi=\ker\varphi=K$. Therefore, by the Fundamental Homomorphism Theorem $H/K\cong H’$.

Suppose that $H’\triangleleft G’$. Let $a\in G$ and $h\in H$. Then $\varphi(a)\in G’$ and $\varphi(h)\in H’$. Since $H’\triangleleft G’$, $\varphi(a)\varphi(h)\varphi(a)^{-1}=\varphi(aha^{-1})\in H’$ which implies that $aha^{-1}\in H$. Hence, $H\triangleleft G$.

Theorem [The Second Isomorphism Theorem]. Let $H\leq G$ and $N\triangleleft G$. Then $HN\leq G$, $H\cap N\triangleleft G$ and
$$H/H\cap N\cong HN/N.$$

Proof. Let $a,b\in HN$. Then $a=h_1n_1$ and $b=h_2n_2$ for some $h_1,h_2\in H$ and $n_1,n_2\in N$. Then
\begin{align*}
ab^{-1}&=h_1n_1(h_2n_2)^{-1}\\
&=h_1n_1(n_2^{-1}h_2^{-1})\\
&=h_1h_2^{-1}h_2(n_1n_2^{-1})h_2^{-1})\in HN.
\end{align*}
So, $HN\leq G$. Clearly $H\cap N\triangleleft G$, in particular $H\cap N\triangleleft H$. Also clearly $N\leq HN$. Let $n_1\in N$. Then $\forall hn\in HN$,
$$(hn)n_1(hn)^{-1}=h(nn_1n^{-1})h^{-1}\in N.$$This means that $N\triangleleft HN$. Define $\varphi:H\longrightarrow HN/N$ by
$$\varphi(h)=Nh$$for each $h\in H$. Then $\varphi$ is clearly well-defined, a homomorphism. Let $Nhn\in HN/N$. Then $Nhn=hnN=hN=Nh$ and $\varphi(h)=Nh=Nhn$. So, $\varphi$ is onto.
\begin{align*}
h\in \ker\varphi&\Leftrightarrow \varphi(h)=Nh=N\\
&\Leftrightarrow h\in N\\
&\Leftrightarrow h\in H\cap N.
\end{align*}
So, $\ker\varphi=H\cap N$. Therefore, by the Fundamental Homomorphism Theorem
$$H/H\cap N\cong HN/N.$$

Theorem [The Third Isomorphism Theorem]. Let $\varphi: G\longrightarrow G’$ be an epimorphism. Let $K=\ker\varphi$, $N’\triangleleft G’$, and $N=\varphi^{-1}(N’)=\{a\in G: \varphi(a)\in N’\}$. Then
$$G/N\cong G’/N,$$or equivalently
$$G/N\cong(G/K)/(N/K).$$

Proof. Define $\psi: G\longrightarrow G’/N’$ by
$$\psi(a)=N’\varphi(a)$$for each $a\in G$. Then

1. $\psi$ is well-defined:\begin{align*}a=b\in G &\Rightarrow\varphi(a)=\varphi(b)\\&\Rightarrow \psi(a)=N’\varphi(a)=N’\varphi(b)=\psi(b).\end{align*}
2. $\psi$ is a homomorphism:\begin{align*}\psi(ab)&=N’\varphi(ab)\\&=N’\varphi(a)\varphi(b)\\&=N’\varphi(a)N’\varphi(b)\\&=\psi(a)\psi(b).\end{align*}
3. $\psi$ is onto: Let $N’c\in G’/N’$. Then $c\in G’$. Since $\varphi$ is onto, there exists $a\in G$ such that $\varphi(a)=c$. $N’c=N’\varphi(a)=\psi(a)$.
4. $\ker\psi=N$:\begin{align*}a\in\ker\psi&\Leftrightarrow N’\varphi(a)=N’\\&\Leftrightarrow\varphi(a)\in N’\\&\Leftrightarrow a\in N.\end{align*}

Therefore, by the Fundamental Homomorphism Theorem we obtain
$$G/N\cong G’/N’.$$Since $\varphi:G\longrightarrow G’$ is an epimorphism, by the Fundamental Homomorphism Theorem, $G’\cong G/K$. Since $N=\varphi^{-1}(N’)$, by the Correspondence Theorem, $N’\cong N/K$. Hence, $G’/N’\cong(G/K)/(N/K)$ i.e. $G/N\cong(G/K)/(N/K)$.

# Functional Analysis 10: Linear Functionals

Definition. Let $X$ be a vector space. A linear functional is a linear map $f:\mathcal{D}(f)\subset X\longrightarrow\mathbb{R}$ (or $f:\mathcal{D}(f)\subset X\longrightarrow\mathbb{C}$).

Definition. A linear functional $f:\mathcal{D}(f)\longrightarrow\mathbb{R}$ is said to be bounded if there exists a number $c$ such that $|f(x)|\leq c||x||$ for all $x\in\mathcal{D}(f)$. Just as in linear operators case $||f||$ is defined by
\begin{align*}
||f||&=\sup_{\begin{array}{c}x\in\mathcal{D}(f)\\x\ne O\end{array}}\frac{|f(x)|}{||x||}\\
&=\sup_{\begin{array}{c}x\in\mathcal{D}(f)\\||x||=1\end{array}}|f(x)|.
\end{align*}
Also we have the inequality holds
$$|f(x)|\leq ||f||||x||$$
for all $x\in\mathcal{D}(f)$.

Just as in linear operators case, we have the following theorem holds.

Theorem. A linear functional $f$ with domain $\mathcal{D}(f)$ in a normed space is continuous if and only if $f$ is bounded.

Example. Let $a=(\alpha_j)\in\mathbb{R}^3$. Define $f:\mathbb{R}^3\longrightarrow\mathbb{R}$ by
$$f(x)=x\cdot a=\xi_1\alpha_1+\xi_2\alpha_2+\xi_3\alpha_3$$ for each $x=(\xi_j)\in\mathbb{R}^3$. Then $f$ is a linear functional. By Cauchy-Schwarz inequality, we obtain
$$|f(x)|=|x\cdot a|\leq ||x||||a||$$
which implies $||f||\leq ||a||$. On the other hand, for $x=a$
$$||a||=\frac{||a||^2}{||a||}=\frac{|f(a)|}{||a||}\leq ||f||.$$
Hence, we have $||f||=||a||$.

Example. Define $f:\mathcal{C}[a,b]\longrightarrow\mathbb{R}$ by
$$f(x)=\int_a^b x(t)dt$$
for each $x(t)\in\mathcal{C}[a,b]$. Then $f$ is a linear functional.
\begin{align*}
|f(x)|&\leq\left|\int_a^b x(t)dt\right|\\
&\leq(b-a)\max|x(t)|\\
&=(b-a)||x||.
\end{align*}
So, $||f||\leq b-a$. Let $x=x_0=1$. Then
\begin{align*}
b-a&=\int_a^b dt\\
&=\frac{|f(x_0)|}{||x_0||}\\
&\leq ||f||.
\end{align*}
Hence, we have $||f||=b-a$.

Let $X^\ast$ be the set of all linear functionals. Then $X\ast$ can be made into a vector space. For any $f,g\in X^\ast$ and scalar $\alpha$, define addition $f+g$ and scalar multiplication $\alpha f$ as follows: For each $x\in X$,
\begin{align*}
(f+g)(x)&=f(x)+g(x),\\
(\alpha f)(x)&=\alpha f(x).
\end{align*}
$X^\ast$ is called the dual space of $X$. One may also consider $X^{\ast\ast}=(X^\ast)^\ast$, the dual space of $X^\ast$. Fix $x\in X$. Define a map $g_x: X^\ast\longrightarrow\mathbb{R}$ by
$$g_x(f)=f(x)$$
for each $f\in X^\ast$. For any $f_1,f_2\in X^\ast$,
\begin{align*}
f_1=f_2&\Longrightarrow f_1(x)=f_2(x)\\
&\Longrightarrow g_x(f_1)=g_x(f_2).
\end{align*}
so, $g_x$ is well-defined. Furthermore, $g_x$ is linear. To show this, for any $f_1,f_2\in X^\ast$ and scalars $\alpha,\beta$,
\begin{align*}
g_x(\alpha f_1+\beta f_2)&=(\alpha f_1+\beta f_2)(x)\\
&=\alpha f_1(x)+\beta f_2(x)\\
&=\alpha g_x(f_1)+\beta g_x(f_2).
\end{align*}
Define a map $C: X\longrightarrow X^{\ast\ast}$ by
$$Cx=g_x$$
for each $x\in X$. Then $C$ is a linear map. First let $x=y\in X$. Then for any $f\in X^\ast$, $g_x(f)=f(x)=f(y)=g_y(f)$, so $C(x)=g_x=g_y=C(y)$. Hence, $C$ is well-defined. To show that $C$ is linear, let $x,y\in X$ and $\alpha,\beta$ scalars. For any $f\in X^\ast$,
\begin{align*}
g_{\alpha x+\beta y}(f)&=f(\alpha x+\beta y)\\
&=\alpha f(x)+\beta f(y)\ (f\ \mbox{is linear})\\
&=\alpha g_x(f)+\beta g_y(f)\\
&=(\alpha g_x+\beta g_y)(f).
\end{align*}
Thus,
$$C(\alpha x+\beta y)=g_{\alpha x+\beta y}=\alpha g_x+\beta g_y=\alpha Cx+\beta Cy.$$
If $X$ is an inner product space or $X$ is a finite dimensional vector space, $C$ becomes oen-to-one. Let us assume that $X$ is equipped with an inner product $\langle\ ,\ \rangle$. Then for any fixed $a\in X$, the map $f_a: X\longrightarrow\mathbb{R}$ defined by
$$f_a(x)=\langle a,x\rangle\ \mbox{for each}\ x\in X$$
is a linear functional. Let $Cx=Cy$. Then $g_{x-y}=0$ and so $g_{x-y}(f_{x-y})=||x-y||^2=0$, hence $x=y$. Therefore, $C$ is one-to-one. We will discussed the case when $X$ is finite dimensional in the next lecture. If $C$ is one-to-one, $X$ is embedded into $X^{\ast\ast}$. We call $C:X\hookrightarrow X^{\ast\ast}$ the canonical embedding. (Here, the notation $\hookrightarrow$ means an embedding or a monomorphism.) If in addition $C$ is onto i.e. $X\stackrel{C}{\cong}X^{\ast\ast}$, then $X$ is said to be algebraically reflexive. If $X$ is finite dimensional, then $X$ is algebraically reflexive. This will be discussed in the next lecture as well.

# Functional Analysis 9: Bounded and Continuous Linear Operators

Definition. Let $X,Y$ be normed spaces and $T:\mathcal{D}(T)\longrightarrow Y$ be a linear operator where $\mathcal{D}(T)\subset X$. $T$ is said to be bounded if there exists $c\in\mathbb{R}$ such that for any $x\in\mathcal{D}(T)$,
$$||Tx||\leq c||x||.$$

Suppose that $x\ne O$. Then
$$\frac{||Tx||}{||x||}\leq c.$$
Let
$$||T||:=\sup_{\begin{array}{c}x\in\mathcal{D}(T)\\ x\ne O\end{array}}\frac{||Tx||}{||x||}.$$
Then $||T||$ is called the norm of the operator $T$. If $\mathcal{D}(T)=\{O\}$ then we define $||T||=0$.

Lemma. Let $T$ be a bounded linear operator. Then

1. $||T||=\displaystyle\sup_{\begin{array}{c}x\in\mathcal{D}(T)\\||x||=1\end{array}}||Tx||.$
2. $||\cdot||$ defined on bounded linear operators satisfies (N1)-(N3).

Proof.

1. \begin{align*}||T||&=\sup_{\begin{array}{c}x\in\mathcal{D}(T)\\x\ne O\end{array}}\frac{||Tx||}{||x||}\frac{||Tx||}{||x||}\\&=\sup_{\begin{array}{c}x\in\mathcal{D}(T)\\x\ne O\end{array}}\left|\left|\frac{||Tx||}{||x||}\right|\right|\\&=\sup_{\begin{array}{c}y\in\mathcal{D}(T)\\||y||=1\end{array}}||Ty||.\end{align*}
2. \begin{align*}||T||=0&\Longleftrightarrow Tx=0,\ \forall x\in\mathcal{D}(T)\\&\Longleftrightarrow T=0.\end{align*} Since $$\sup_{\begin{array}{c}x\in\mathcal{D}(T)\\||x||=1\end{array}}||(T_1+T_2)x||\leq \sup_{\begin{array}{c}x\in\mathcal{D}(T)\\||x||=1\end{array}}||T_1x||+\sup_{\begin{array}{c}x\in\mathcal{D}(T)\\||x||=1\end{array}}||T_2x||,$$ $$||T_1+T_2||\leq ||T_1||+||T_2||.$$

Examples.

1. The identity operator $I:X\longrightarrow X$ with $X\ne\{O\}$ is a bounded linear operator with $||I||=1$.
2. Zero operator $O: X\longrightarrow Y$ is a bounded linear operator with $||O||=0$.
3. Let $X$ be the normed space of all polynomials on $[0,1]$ with $||x||=\max_{t\in[0,1]}|x(t)|$. Differentiation$$T: X\longrightarrow X;\ Tx(t)=x’(t)$$ is not a bounded operator. To see this, let $x_n(t)=t^n$, $n\in\mathbb{N}$. Then $||x_n||=1$ for all $n\in\mathbb{N}$. $Tx_n(t)=nt^{n-1}$ and $||Tx_n||=n$, for all $n\in\mathbb{N}$. So, $\frac{||Tx_n||}{||x_n||}=n$ and hence $||T||$ is not bounded.
4. Integral operator $$T:\mathcal{C}[0,1]\longrightarrow\mathcal{C}[0,1];\ Tx=\int_0^1\kappa(t,\tau)x(\tau)d\tau$$ is a bounded linear operator. The function $\kappa(t,\tau)$ is a continuous function on $[0,1]\times[0,1]$ called the kernel of $T$. \begin{align*}||Tx||&=\max_{t\in[0,1]}\left|\int_0^1\kappa(t,\tau)x(\tau)d\tau\right|\\&\leq\max_{t\in[0,1]}\int_0^1|\kappa(t,\tau)||x(\tau)|d\tau\\&\leq k_0||x||,\end{align*}where $k_0=\displaystyle\max_{(t,\tau)\in[0,1]\times[0,1]}\kappa(t,\tau)$.
5. Let $A=(\alpha_{jk})$ be an $r\times n$ matrix of real entries. The linear map $T:\mathbb{R}^n\longrightarrow\mathbb{R}^r$ given by $Tx=Ax$ for each $x\in\mathbb{R}^n$ is bounded. To see this, Let $x\in\mathbb{R}^n$ and write $x=(\xi_j)$. Then $||x||=\sqrt{\displaystyle\sum_{m=1}^n\xi_m^2}$.\begin{align*}||Tx||^2&=\sum_{j=1}^r\left[\sum_{k=1}^n\alpha_{jk}\xi_k\right]^2\\&\leq\sum_{j=1}^r\left[\left(\sum_{k=1}^n\alpha_{jk}^2\right)^\frac{1}{2}\left(\sum_{m=1}^n\xi_m\right)^\frac{1}{2}\right]^2\\&=||x||^2\sum_{j=1}^r\sum_{k=1}^n\alpha_{jk}^2.\end{align*}By setting $c^2=\displaystyle\sum_{j=1}^r\sum_{k=1}^n\alpha_{jk}^2$, we obtain$$||Tx||^2\leq c^2||x||^2.$$

In general, if a normed space $X$ is finite dimensional, then every linear operator on $X$ is bounded. Before we discuss this, we first introduce the following lemma without proof.

Lemma. Let $\{x_1,\cdots,x_n\}$ be a linearly independent set of vectors in a normed space $X$. Then there exist a number $c>0$ such that for any scalars $\alpha_1,\cdots,\alpha_n$, we have the inequality
$$||\alpha_1x_1+\cdots+\alpha_nx_n||\geq c(|\alpha_1|+\cdots+|\alpha_n|).$$

Theorem. If a normed space $X$ is finite dimensional, then every linear operator on $X$ is bounded.

Proof. Let $\dim X=n$ and $\{e_1,\cdots,e_n\}$ be a basis for $X$. Let $x=\displaystyle\sum_{j=1}^n\xi_je_j\in X$. Then
\begin{align*}
||Tx||&=||\sum_{j=1}^n\xi_jTe_j||\\
&\leq\sum_{j=1}^n||\xi_j|||Te_j||\\
&\leq\max_{k=1,\cdots,n}||Te_k||\sum_{j=1}^n|\xi_j|.
\end{align*}
By Lemma, there exists a number $c>0$ such that
$$||x||=||\xi_1e_1+\cdots+\xi_ne_n||\geq c(|\xi_1|+\cdots+|\xi_n|)=c\sum_{j=1}^n|\xi_j|.$$
So, $\displaystyle\sum_{j=1}^n|\xi_j|\leq\frac{1}{c}||x||$ and hence
$$||Tx||\leq M||x||,$$ where
$M=\frac{1}{c}\max_{k=1,\cdots,n}||Te_k||$.

What is really nice about linear operators from a normed space into a normed space is that a linear operator being bounded is equivalent to it being continuous.

Theorem. Let $X,Y$ be normed spaces and $T:\mathcal{D}(T)\subset X\longrightarrow Y$ a linear operator. Then

1. $T$ is continuous if and only if $T$ is bounded.
2. If $T$ is continuous at a single point, it is continuous.

Proof.

1. If $T=O$, then we are done. Suppose that $T\ne O$. Then $||T||\ne 0$. Assume that $T$ is bounded and $x_0\in\mathcal{D}(T)$. Let $\epsilon>0$ be given. Choose $\delta=\frac{\epsilon}{||T||}$. Then for any $x\in\mathcal{D}(T)$ such that $||x-x_0||<\delta$, $$||Tx-Tx_0||=||T(x-x_0)||\leq ||T||||x-x_0||<\epsilon.$$ Conversely, assume that $T$ is continuous at $x_0\in\mathcal{D}(T)$. Then given $\epsilon>0$ there exists $\delta>0$ such that $||Tx-Tx_0||<\epsilon$ whenever $||x-x_0||\leq\delta$. Take $y\ne 0\in\mathcal{D}(T)$ and set $$x=x_0+\frac{\delta}{||y||}y.$$ Then $x-x_0=\frac{\delta}{||y||}y$ and $||x-x_0||=\delta$. So,\begin{align*}||Tx-Tx_0||&=||T(x-x_0)||\\&=\left|\left|T\left(\frac{\delta}{||y||}y\right)\right|\right|\\&=\frac{\delta}{||y||}Ty\\&<\epsilon.\end{align*}Hence, for any $y\in\mathcal{D}(T)$, $||Ty||\leq\frac{\epsilon}{\delta}||y||$ i.e. $T$ is bounded.
2. In the proof of part (a), we have shown that if $T$ is continuous at a point, it is bounded. If $T$ is bounded, then it is continuous by part (a).

Corollary. Let $T$ be a bounded linear operator. Then

1. If $x_n\to x$ then $Tx_n\to Tx$.
2. $\mathcal{N}(T)$ is closed.

Proof.

1. If $T$ is bounded, it is continuous and so the statement is true.
2. Let $x\in\overline{\mathcal{N}(T)}$. Then there exists a sequence $(x_n)\subset\mathcal{N}(T)$ such that $x_n\to x$. Since $Tx_n=0$ for each $n=1,2,\cdots$, $Tx=0$. Hence, $x\in\mathcal{N}(T)$.

Theorem. Let $X$ be a normed space and $Y$ a Banach space. Let $T:\mathcal{D}(T)\subset X\longrightarrow Y$ be a bounded linear operator. Then $T$ has an extension $\tilde T:\overline{\mathcal{D}(T)}\longrightarrow Y$ where $\tilde T$ is a bounded linear operator of norm $||\tilde T||=||T||$.

Proof. Let $x\in\overline{\mathcal{D}(T)}$. Then there exists a sequence $(x_n)\subset\mathcal{D}(T)$ such that $x_n\to x$. Since $T$ is bounded and linear,
\begin{align*}
||Tx_m-Tx_n||&=||T(x_m-x_n)||\\
&\leq||T||||x_m-x_n||,
\end{align*}
for all $m,n\in\mathbb{N}$. Since $(x_n)$ is convergent, it is Cauchy so given $\epsilon>0$ there exists a positive integer $N$ such that for all $m,n\geq N$, $||x_m-x_n||<\frac{\epsilon}{||T||}$. Hence, for all $m,n>N$,
\begin{align*}
||Tx_m-Tx_n||&\leq ||T||||x_m-x_n||\\
&<\epsilon.
\end{align*}
That is, $(Tx_n)$ is a Cauchy sequence in $Y$. Since $Y$ is a Banach space, there exists $y\in Y$ such that $Tx_n\to y$. Define $\tilde T:\overline{\mathcal{D}(T)}\longrightarrow Y$ by $\tilde Tx=y$. In order for $\tilde T$ to be well- defined, its definition should not depend on the choice $(x_n)$. Suppose that there is a sequence $(z_n)\subset\mathcal{D}(T)$ such that $z_n\to x$. Then $x_n-z_n\to 0$. Since $T$ is bounded, it is continuous so $T(x_n-z_n)\to 0$. This means that $\displaystyle\lim_{n\to\infty}Tz_n=\lim_{n\to\infty}Tx_n=y$. $\tilde T$ is linear and $\tilde T|_{\mathcal{D}(T)}=T$. To show that $\tilde T$ is bounded, let $x\in\overline{\mathcal{D}(T)}$. Then there exists a sequence $(x_n)\subset\mathcal{D}(T)$ such that $x_n\to x$ as before. Since $T$ is bounded, for each $n=1,2,\cdots$,
$$||Tx_n||\leq ||T||||x_n||.$$ Since the norm $x\longmapsto||x||$ is continuous, as $n\to\infty$ we obtain
$$||\tilde Tx||\leq ||T||||x||.$$ Hence, $\tilde T$ is bounded and $||\tilde T||\leq ||T||$. On the other hand, since $\tilde T$ is an extension of $T$, $||T||\leq||\tilde T||$. Therefore, $||\tilde T||=||T||$.

# Functional Analysis 8: Linear Operators

From here on, a map from a vector space into another vector space will be called an operator.

Definition. A linear operator $T$ is an operator such that

1. $T(x+y)=Tx+Ty$ for any two vectors $x$ and $y$.
2. $T(\alpha x)=\alpha Tx$ for any vector $x$ and a scalar $\alpha$.

Proposition. An operator $T$ is a linear operator if and only if
$$T(\alpha x+\beta y)=\alpha Tx+\beta Ty$$
for any vectors $x,y$ and scalars $\alpha,\beta$.

Denote by $\mathcal{D}(T)$, $\mathcal{R}(T)$ and $\mathcal{N}(T)$, the domain, the range and the null space, respectively, of a linear operator $T$. The null space $\mathcal{N}(T)$ is the kernel of $T$ i.e.
$$\mathcal{N}(T)=T^{-1}(0)=\{x\in \mathcal{D}(T): Tx=0\}.$$
Since the term kernel is reserved for something else in functional analysis, we call it the null space of $T$.

Example. [Differentiation] Let $X$ be the space of all polynomials on $[a,b]$. Define an operator $T: X\longrightarrow X$ by
$$Tx(t)=x’(t)$$
for each $x(t)\in X$. Then $T$ is linear and onto.

Example. [Integration] Recall that $\mathcal{C}[a,b]$ denotes the space of all continuous functions on the closed interval $[a,b]$. Define an operator $T:\mathcal{C}[a,b]\longrightarrow\mathcal{C}[a,b]$ by
$$Tx(t)=\int_a^tx(\tau)d\tau$$
for each $x(t)\in\mathcal{C}[a,b]$. Then $T$ is linear.

Example. Let $A=(a_{jk})$ be an $r\times n$ matrix of real entries. Define an operator $T: \mathbb{R}^n\longrightarrow\mathbb{R}^r$ by
$$Tx=Ax=(a_{jk})(\xi_l)=\left(\sum_{k=1}^na_{jk}\xi_k\right)$$
for each $n\times 1$ column vector $x=(\xi_l)\in\mathbb{R}^n$. Then $T$ is linear as seen in linear algebra.

Theorem. Let $T$ be a linear operators. Then

1. The range $\mathcal{R}(T)$ is a vector space.
2. If $\dim\mathcal{D}(T)=n<\infty$, then $\dim\mathcal{R}\leq n$.
3. The null space $\mathcal{N}(T)$ is a vector space.

Proof. Parts 1 and 3 are straightforward. We prove part 2. Choose $y_1,\cdots,y_{n+1}\in\mathcal{R}(T)$. Then $y_1=Tx_1,\cdots,y_{n+1}=Tx_{n+1}$ for some $x_1,\cdots,\\x_{n+1}\in\mathcal{D}(T)$. Since $\dim\mathcal{D}(T)=n$, $x_1,\cdots,x_{n+1}$ are linearly dependent. So, there exist scalars $\alpha_1,\cdots,\alpha_{n+1}$ not all equal to 0 such that $\alpha_1x_1+\cdots+\alpha_{n+1}x_{n+1}=0$. Since $T(\alpha_1x_1+\cdots+\alpha_{n+1}x_{n+1})=\alpha_1y_1+\cdots+\alpha_{n+1}y_{n+1}=0$, $\mathcal{R}$ has no linearly independent subset of $n+1$ or more elements.

Theorem. $T$ is one-to-one if and only if $\mathcal{N}=\{O\}$.

Proof. Suppose that $T$ is one-to-one. Let $a\in\mathcal{N}$. Then $Ta=O=TO$. Since $T$ is one-to-one, $a=O$ and hence $\mathcal{N}=\{O\}$. Suppose that $\mathcal{N}=\{O\}$. Let $Ta=Tb$. Then by linearity $T(a-b)=O$ and so $a-b\in\mathcal{N}=\{O\}\Longrightarrow a=b$. Thus, $T$ is one-to-one.
Theorem.

1. $T^{-1}: \mathcal{R}(T)\longrightarrow\mathcal{D}(T)$ exists if and only if $\mathcal{N}=\{O\}$ if and only if $T$ is one-to-one.
2. If $T^{-1}$ exists, it is linear.
3. If $\dim\mathcal{D}(T)=n<\infty$ and $T^{-1}$ exists, then $\dim\mathcal{R}(T)=\dim\mathcal{D}(T)$.

Proof. Part 1 is trivial. Part 3 follows from part 2 of the previous theorem. Let us prove part 2. Let $y_1,y_2\in\mathcal{R}(T)$. Then there exist $x_1,x_2\in\mathcal{D}(T)$ such that $y_1=Tx_1$, $y_2=Tx_2$. Now,
\begin{align*}
\alpha y_1+\beta y_2&=\alpha Tx_1+\beta Tx_2\\
&=T(\alpha x_1+\beta x_2).
\end{align*}
So,
\begin{align*}
T^{-1}(\alpha y_1+\beta y_2)&=T^{-1}(T(\alpha x_1+\beta x_2))\\
&=\alpha x_1+\beta x_2\\
&=\alpha T^{-1}y_1+\beta T^{-1}y_2.
\end{align*}

# 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.

# Group Theory 10: 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, $\mathrm{SL}(2,\mathbb{R})\vartriangleleft\mathrm{GL}(2,\mathbb{R})$.