# Abstract Nonsense

## Review of Group Theory: The Structure Theorem For Finite Abelian Groups[Almost](Pt. I)

Point of post: In this post we discuss an prove a weak version of the structure theorem for finite abelian groups. Namely, for any abelian group $G$ there exists primes $p_1,\cdots,p_m$ and integers $n_1,\cdots,n_m$ such that $G\cong\mathbb{Z}_{p_1^{n_1}}\times\cdots\times\mathbb{Z}_{p_m^{n_m}}$ and that this representation is unique.

Motivation

It’s always been amazing how much simpler the theory for finite abelian groups has been then that of non-abelian groups. I mean, ostensibly the commutativity of the binary operation does seem like it would make things simpler, but really the theory of non-abelian groups is on another whole level. In this post we shall quantify how big of a difference it is by showing that every abelian group is just a direct product of cyclic groups whose orders are powers of primes. And that up to permutation of product factors this decomposition is unique. The difference between this and the case for non-abelian groups. In fact, the only similar theorem which says roughly “you only need to study this family of groups because the rest of the theory you care about is secretly hidden within” is Cayley’s theorem which makes the ‘parent’ family of groups the symmetric groups–which are not by any standard as nice as the direct product of cyclic groups whose orders are powers of primes.

Remark: The almost part is because I will only prove the existence part of the proof and not the uniqueness. The reason for this is that the uniqueness proof is my interpretation of a clever proof of this theorem given in Barry Simon’s book. The uniqueness part he gives I am not nearly as fond of. And, since I know a proof and it is the proof that is in all the books (Rotman, Dummit and Foote, Lang, Grillet, etc.) there is no use copying it here.

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Structure Theorem for Finite Abelian Groups

We begin a lemma which can be seen as a corollary from a previous theorem:

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Lemma 1: Let $n=a_1\cdots a_m$ where $(a_1,\cdots,a_m)=1$. Then,

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$\mathbb{Z}_n\cong\mathbb{Z}_{a_1}\times\cdots\times\mathbb{Z}_{a_m}$

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Proof: This follow immediately from our previous theorem that if $(a,b)=1$ then $\mathbb{Z}_{ab}\cong\mathbb{Z}_a\times\mathbb{Z}_b$ applied multiple times. $\blacksquare$

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Corollary: If $p_1,\cdots,p_m$ are primes and $n=p_1^{n_1}\cdots p_m^{n_m}$ then:

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$\mathbb{Z}_{n}\cong\mathbb{Z}_{p_1^{n_1}}\times\cdots\times\mathbb{Z}_{p_m^{n_m}}$

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We next show then that any abelian group $G$ is isomorphic to a group of the form $\mathbb{Z}_{a_1}\times\cdots\mathbb{Z}_{a_m}$:

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Lemma 2: Let $G$ be a finite abelian group, then there exists $a_1,\cdots,a_m\in\mathbb{N}$ such that

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$G\cong\mathbb{Z}_{a_1}\times\cdots\times\mathbb{Z}_{a_m}$

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Proof: We say that a group is $r$-generatable if there exists $\{g_1,\cdots,g_r\}\subseteq G$ such that $\{g_1,\cdots,g_r\}$ generates. Clearly every finite group $G$ is $|G|$-generatable since $\langle\{g\}_{g\in G}\rangle=G$. We induct on $r$. For $r=1$ this is trivial since by definition every $1$-generatable group $G$  is cyclic and thus isomorphic to $\mathbb{Z}_{|G|}$.

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Suppose then that every $r-1$-generatable group can be written as the direct product of cyclic groups and let $G$ be $r$-generatable. For any $m\in\mathbb{N}$ and any $m$-tuple $(a_1\cdots,a_m)\subseteq G^r$ define $f_{(a_1,\cdots,a_m)}:\mathbb{Z}^m\to G$ by

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$\displaystyle (n_1,\cdots,n_m)\mapsto \sum_{j=1}^{m}n_j a_j$

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It’s clear via the fact that $G$ is abelian that $f_{(a_1,\cdots,a_m)}$ is a homomorphism. Let then $R_{(a_1,\cdots,a_m)}=\ker f_{(a_1,\cdots,a_m)}$. If we assume that $(a_1,\cdots,a_m)$ generates $G$ then by definition $f_{(a_1,\cdots,a_m)}$ is an epimorphism and so by the first isomorphism theorem we have $\mathbb{Z}^r/R_{(a_1,\cdots,a_m)}\cong G$. Now, let $\pi:\mathbb{Z}^r\to\mathbb{Z}$ be the first canonical projection onto the first coordinate given by $(n_1,\cdots,n_r)\mapsto n_1$ and define

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$\displaystyle S=\mathbb{N}\cap\bigcup_{\substack{(a_1,\cdots,a_r)\text{ which}\\ \text{generate }G}}\pi\left(R_{(a_1,\cdots,a_r)}\right)$

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and let $\widetilde{n_1}=\min S$ and find $(\widetilde{a_1},\cdots,\widetilde{a_r})$ which generates $G$ such that $(\widetilde{n_1},\cdots,\widetilde{n_r})\in R_{(\widetilde{a_1},\cdots,\widetilde{a_r})}$ for some $\widetilde{n_2},\cdots,\widetilde{n_r}\in\mathbb{Z}$. We now claim that if $(n_1,\cdots,n_r)\in R_{(\widetilde{a_1},\cdots,\widetilde{a_r})}$ then $\widetilde{n_1}\mid n_1$. Indeed, by the division algorithm we have that $n_1=q\widetilde{n_1}+m$ where $0\leqslant m<\widetilde{n_1}$ and

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$\displaystyle 0=\sum_{j=1}^{r}n_j\widetilde{a_j}-q\sum_{j=1}^{r}q\widetilde{n_j}\widetilde{a_j}=ma_1+\sum_{j=2}^{r}(n_j-q\widetilde{n_j})\widetilde{a_j}$

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and so $(m,\widetilde{n_2}-q\widetilde{n_2},\cdots,\widetilde{n_r}-q\widetilde{n_r})\in R_{(\widetilde{a_1},\cdots,\widetilde{a_r})}$ and thus by construction we must have that $m=0$ from where $\widetilde{n_1}\mid n_1$ follows. It follows then that if $(n_1,n_2,\cdots,n_r)\in R_{(\widetilde{a_1},\cdots,\widetilde{a_r})}$ that $n_1=q\widetilde{n_1}$ for some $q\in\mathbb{Z}$ and so

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$(n_1,n_2,\cdots,n_r)=q(\widetilde{n_1},,\widetilde{n_2},\cdots,\widetilde{n_r})+(0,n_2-q\widetilde{n_2},\cdots,n_r-q\widetilde{n_r})\quad\mathbf{(1)}$

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but it’s easy to show that $(n_2-q\widetilde{n_2},\cdots,n_r-q\widetilde{n_r})\in R_{(\widetilde{a_2},\cdots,\widetilde{a_r})}$. What we claim though next is that $\widetilde{n_1}\mid \widetilde{n_k}$ for $k=2,\cdots,r$. Indeed, by the division algorithm write $\widetilde{n_k}=q_k\widetilde{n_1}+m_k,\;0\leqslant m_k<\widetilde{n_1}$ for $k=2,\cdots,r$ and let $\displaystyle \widetilde{a}=a_1+q_2\widetilde{a_2}+\cdots+q_r\widetilde{a_r}$. Clearly then $\{\widetilde{a},\widetilde{a_2},\cdots,\widetilde{a_r}\}$ generate $G$ and since $\widetilde{n_1}\widetilde{a}+m_2\widetilde{a_2}+\cdots+m_r\widetilde{a_r}=0$ we must have $m_k=0$ for $k=2,\cdots,r$ otherwise if, say $m_{k_0}\ne 0$, then $\left(m_{k_0},\widetilde{n_1},\cdots\right)\in R_{(\widetilde{a},\widetilde{a_2},\cdots,\widetilde{a_r})}$ and $m_{k_0}<\widetilde{n_1}$ contradicting the minimality of $\widetilde{n_1}$. Thus, $(\widetilde{n_1},\widetilde{n_2},\cdots,\widetilde{n_r})=\widetilde{n_1}(1,q_2,\cdots,q_r)$. Note though that if we define $A=\left\{q(\widetilde{n_1},\cdots,\widetilde{n_r}):q\in\mathbb{Z}\right\}$ and $B=\left\{(0,n_2,\cdots,n_r):(n_2,\cdots,n_r)\in R_{(\widetilde{a_2},\cdots,\widetilde{a_r})}\right\}$ then by $\mathbf{(1)}$ we have already proven that $R_{(\widetilde{a_1},\cdots,\widetilde{a_r})}=A+B$. Noting then that $A\cap B$ is trivial it easily follows that $R_{(\widetilde{a_1},\cdots,\widetilde{a_r})}\cong A\times B$. Note though that by prior observation we have that $A=\left\{q\widetilde{n_1}(1,q_1,\cdots,q_r):q\in\mathbb{Z}\right\}$ and so evidently $A\cong \widetilde{n_1}\mathbb{Z}$. Moreover, it’s clear that $B\cong R_{(\widetilde{a_2},\cdots,\widetilde{a_r})}$ and that if $H=\text{im}f_{(\widetilde{a_2},\cdots,\widetilde{a_r})}=\left\langle \widetilde{a_2},\cdots,\widetilde{a_r}\right\rangle$ then $\mathbb{Z}^{r-1}/R_{(\widetilde{a_2},\cdots,\widetilde{a_r})}\cong H$. Thus

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$G\cong \mathbb{Z}^r/R_{(\widetilde{a_1},\cdots,\widetilde{a_r})}\cong \left(\mathbb{Z}/\widetilde{n_1}\mathbb{Z}\right)\times \left(\mathbb{Z}^{r-1}/R_{(\widetilde{a_2},\cdots,\widetilde{a_r})}\right)\cong \mathbb{Z}_{\widetilde{n_1}}\times H\quad\mathbf{(2)}$

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But since $H$ is, by construction, $r-1$-generatable it is by the induction hypothesis isomorphic to a direct product of cyclic groups and so evidently from $\mathbf{(2)}$ so is $G$. The induction is complete. $\blacksquare$

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References:

1. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Math. Soc., 1996. Print.

April 16, 2011 -

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