Monthly Archives: December 2014

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. $360=2^3\cdot 3^2\cdot 5$, so all abelian groups, up to isomorphism, of order 360 are

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
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
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
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

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.