Group Theory

Consider an abelian group $A$ generated by $m$ elements

$$A=⟨a_1,...,a_m⟩$$

Then the free abelian group of rank $m$

$$F=⟨u_1,...,u_m⟩$$

maps homomorphically onto $A$ via the map that sends $u_i$ to $a_i$. By the first isomorphism theorem we have $A\cong F/R$ for some subgroup $R$ of $F$. Pick a basis $v_1,...,v_m$ of $F$ such that $R=⟨h_1{v}_1,...,h_q{v}_q⟩$ where $h_i\mid h_{i+1}$, $h_i\ge 1$, $q\le m$.

Consider the case where $m=1$. There are three possibilities. (1) $R=⟨v⟩$, so $F/R$ is the trivial group, (2) $R=⟨hv⟩$, in which case $F/R={\mathbb{Z}}_h$, and (3) $R=\{0\}$ and we have $F/R=F$.

More generally, we have:

Theorem: Every finitely generated abelian group can be expressed as the direct sum of cyclic groups

$$A={\mathbb{Z}}^n\oplus {\mathbb{Z}}_{h_1}\oplus ...\oplus {\mathbb{Z}}_{h_n}$$

where $h_i\mid h_{i+1}$.

Corollary: A finitely generated abelian group is free if and only if it is torsion-free, that is, it contains no element of finite order other than the identity.

The number $r$ is called the rank of $A$. The orders of the cyclic groups $h_1,...,h_n$ are called the invariants of $A$. Note $A$ is finite if and only if its rank is zero.

Theorem: Suppose $A$ is a finitely generated abelian group with decompositions

$$\begin{array}{c}A={\mathbb{Z}}^r\oplus {\mathbb{Z}}_{e_1}\oplus ...\oplus {\mathbb{Z}}_{e_n}\\ A={\mathbb{Z}}^s\oplus {\mathbb{Z}}_{d_1}\oplus ...\oplus {\mathbb{Z}}_{d_m}\end{array}$$

satifying $e_i\mid e_{i+1},d_i\mid d_{i+1}$. Then $r=s,n=m,e_i=d_i$.

Proof: Let $T$ be the set of elements of $A$ of finite order. Clearly if $g,h$ have finite order then $\mathrm{ord}(g)\mathrm{ord}(h)(h-k)=0$ hence $h-k$ also has finite order hence $T$ is a subgroup of $A$. It is called the torsion group of $A$.

A little thought shows that we must have

$$\begin{array}{c}T={\mathbb{Z}}_{e_1}\oplus ...\oplus {\mathbb{Z}}_{e_n}\\ T={\mathbb{Z}}_{d_1}\oplus ...\oplus {\mathbb{Z}}_{d_m}\end{array}$$

Consider the map that projects $A$ onto ${\mathbb{Z}}^r$. By the first isomorphism theorem we have that $A/T\cong {\mathbb{Z}}^r$. Similarly we have $A/T\cong {\mathbb{Z}}^s$ hence $r=s$.

Now conisder $T$. Let $p$ be a prime, and let $P$ be the set of elements whose order is a power of $P$. Then $P$ is a group. We first need the following:

Theorem: Let $G$ be a finite abelian group of order $p_1^{a_1}{p}_2^{a_2}...$ where the $p_i$'s are distinct primes. Then $G=P_1\oplus P_2\oplus ...$ where $P_i$ is the subgroup of elements whose orders are powers of $p_i$.

Proof: Let $x\in G$ be an element of order $p_1^{\alpha }{f}_1$ where $f_1,p_1$ are coprime. Then we may write $x=a_1+x_1$ where $a_1$ has order $p_1^{\alpha }$ and $x_1$ has order $f_1$. (Simply take $a_1=u{f}_{1}x,x_1=v{p}_1^{\alpha }x$ where $u{f}_1+v{p}_1^{\alpha }=1$.)

Iterating this procedure gives a decomposition $x=a_1+a_2+...$ with $a_i\in P_i$. We claim this decomposition is unique. Suppose $0=b_1+b_2+...$ where $b_i\in P_i$. Then for all $i$, subtracting $b_i$ from both sides shows that the order of $b_i$ is coprime to $p_i$. But it must also be a power of $p_i$ which is only possible if $b_i=0$.

It is clear that the groups $P_i$ are uniquely determined. In fact, they are the Sylow groups since $G$ is abelian.$\blacksquare$

In particular, if $x$ is an element of order $n=p_1^{a_1}{p}_2^{a_2}...$ then we have

$$⟨x⟩=⟨(n/p_1^{a_1})x⟩\oplus ⟨(n/p_2^{a_2})x⟩\oplus ...$$

Now let

$$e_1=p_1^{a_1}{p}_2^{a_2}...,e_2=p_1^{b_1}{p}_2^{b_2}...,...$$

where $a_i\le b_i\le ...$ for all $i$ since $e_i\mid e_i+1$. Then we have

$$T=P_1\oplus P_2\oplus ...\oplus Q_1\oplus Q_2\oplus ...$$

where $P_1,P_2,...,Q_1,Q_2,...$ are cyclic groups of order $p_1^{a_1},p_1^{b_1},...,p_2^{a_2},p_2^{b_2},...$. We see that the Sylow groups of $T$ are $P=P_1\oplus P_2\oplus ...,Q=Q_1\oplus Q_2\oplus ...$. Now we need the following:

Lemma: Let $G$ be any group. Suppose $x,y\in G$ commute and have relatively prime orders $m,n$. Then

$$⟨x,y⟩=⟨xy⟩$$

is cyclic of order $mn$.

Proof: We know the order is at most $mn$ since each element must be of the form $x^a{y}^b$ for $a=0,...,m-1,b=0,...,n-1$. Now suppose $(xy{)}^t=1$. Then $1=(xy{)}^{tm}=y^{tm}$ implying that $n\mid tm$. Since $m,n$ is coprime we have $n\mid t$. Similarly $m\mid t$, thus the group order must be exactly $mn$. $\blacksquare$

Thus given

$$T=P_1\oplus P_2\oplus ...\oplus Q_1\oplus Q_2\oplus ...$$

we deduce that

$$T={\mathbb{Z}}_{e_1}\oplus ...\oplus {\mathbb{Z}}_{e_n}$$

so that one decomposition implies the other. We are done as soon as we show that the Sylow groups have a unique decomposition:

Theorem: Let $A$ be an abelian group of order $p^a$ where $p$ is prime. Suppose

$$\begin{array}{ccc}A& =& ⟨u_1⟩\oplus ...\oplus ⟨u_k⟩\\ A& =& ⟨v_1⟩\oplus ...\oplus ⟨v_l⟩\end{array}$$

where $u_1,...,u_k$ have orders $p^{f_1}\ge ...\ge p^{f_k}>1$, and $v_1,...,v_v$ have orders $p^{g_1}\ge ...\ge p^{g_l}>1$. Then $k=l$ and $f_i=g_i$ for $i=1,...,k$.

Proof: Note we must have $a=f_1+...+f_k=g_1+...+g_l$. The theorem is trivial when $a=1$, which we use to start an induction.

Let $A_p$ be the set of elements $x\in A$ with $px=0$. Then $A_p$ is a subgroup. We have

$$\begin{array}{ccc}{A}_p& =& ⟨p^{f_1-1}{u}_1⟩\oplus ...\oplus ⟨p^{f_k-1}{u}_k⟩\\ A_p& =& ⟨p^{g_1-1}{v}_1⟩\oplus ...\oplus ⟨p^{g_l-1}{v}_l⟩\end{array}$$

Hence $A_p={\mathbb{Z}}_p^k={\mathbb{Z}}_p^l$ implying that $k=l$.

Now consider the set $A^p$ of elements $px$ for all $x\in A$ (the multiples of $p$). Then $A^p$ is a subgroup, and is generated by $p{u}_1,...,p{u}_k$ and also by $p{v}_1,...,p{v}_k$. But in general these are not bases for $A^p$ since we might have $p{u}_i=0$ for example. So find $\kappa$ such that $f_1,...,f_{\kappa }\ge 2$ and $f_{\kappa +1}=...=f_k=1$, and similarly find $\lambda$ with $g_1,...,g_{\lambda }\ge 2$ and $g_{\lambda +1}=...=g_k=1$.

This yields the decompositions

$$A^p=⟨p{u}_1⟩+...+⟨p{u}_{\kappa }⟩=⟨p{v}_1⟩+...+⟨p{v}_{\lambda }⟩$$

By inductive hypothesis we have $\kappa =\lambda$ and $f_i-1=g_i-1$ for all $i=1,...,\kappa$. $\blacksquare$

We have now proved the main theorem. $\blacksquare$

In the last proof, the numbers $p^{f_1},...,p^{f_k}$ are called the elementary divisors of $A$ corresponding to $p$. $A$ is said to be of type $(f_1,...,f_k)$.

Example: Suppose an abelian group $A$ is generated by $a,b$ subject to the relations $30a=12b=0$. Then define the free abelian groups $F=⟨x,y⟩$ and $R=⟨30x,12y⟩$. Note we have $A\cong F/R={\mathbb{Z}}_{30}\oplus {\mathbb{Z}}_{12}$. Then we have

$$A\cong {\mathbb{Z}}_2\oplus {\mathbb{Z}}_3\oplus {\mathbb{Z}}_5\oplus {\mathbb{Z}}_4\oplus {\mathbb{Z}}_3\cong ({\mathbb{Z}}_4\oplus {\mathbb{Z}}_2)\oplus ({\mathbb{Z}}_3\oplus {\mathbb{Z}}_3)\oplus {\mathbb{Z}}_5$$

Thus the elementary divisors for 2,3,5 are $(\mathrm{4,2}),(\mathrm{3,3}),5$. Rearranging gives $A\cong {\mathbb{Z}}_{60}\oplus {\mathbb{Z}}_6$, so the invariants are $60,6$.

Example: Suppose an abelian group $A$ is generated by $a,b,c,d$ and the relations $3a+9b-3c=0,4a+2b-2d=0$. Then define the free abelian groups $F=⟨x,y,z,t⟩$ and $R=⟨3u\mathrm{,2}v⟩$ where $u=x+3y-z,v=2x+y-t$. Note $x,y,u,v$ is also a basis of $F$. Thus

$$A\cong F/R\cong \mathbb{Z}\oplus \mathbb{Z}\oplus {\mathbb{Z}}_3\oplus {\mathbb{Z}}_2\cong \mathbb{Z}\oplus \mathbb{Z}\oplus {\mathbb{Z}}_6$$