Abstract Nonsense

Crushing one theorem at a time

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.


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.

\text{ }

Structure Theorem for Finite Abelian Groups

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

\text{ }

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

\text{ }


\text{ }

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

\text{ }

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

\text{ }


\text{ }

\text{ }

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}:

\text{ }

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

\text{ }


\text{ }

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|}.

\text{ }

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

\text{ }

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

\text{ }

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

\text{ }

\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)

\text{ }

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

\text{ }

\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}

\text{ }

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

\text{ }


\text{ }

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

\text{ }

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)}

\text{ }

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

\text{ }

\text{ }


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


April 16, 2011 - Posted by | Algebra, Group Theory | , , , ,


  1. […] between cyclic groups and their duals, and products of groups and their duals. But, we also know that abelian groups and products of cyclic groups are intimately related. Thus, putting these […]

    Pingback by Representation Theory: The Dual Group of an Abelian Group « Abstract Nonsense | April 16, 2011 | Reply

  2. […] as we were able to combine the structure theorem and our knowledge of the dual group of a cyclic group to gain information about the dual group of […]

    Pingback by Representation Theory: Irreps of an Abelian Group « Abstract Nonsense | April 16, 2011 | Reply

  3. […] as stated we know that every group of order is abelian and, so from the structure theorem we may conclude that if with prime then […]

    Pingback by Groups of Order pq (pt. I) « Abstract Nonsense | April 19, 2011 | Reply

  4. […] that is inevitably cyclic and so by a common theorem we must have that is abelian. Then, by the structure theorem we may conclude that either or […]

    Pingback by University of Maryland College Park Qualifying Exams (Group Theory and Representation Theory) ( January 2003)) « Abstract Nonsense | May 1, 2011 | Reply

  5. […] We know by the structure theorem that for some and some . Thus, it suffices to prove this for groups of that form. We begin by […]

    Pingback by University of Maryland College Park Qualifying Exams (Group Theory and Representation Theory) (January-2004) (Pt. I) « Abstract Nonsense | May 6, 2011 | Reply

  6. […] Really, if is a finite ring then we know from the structure theorem for finite abelian groups that is isomorphic (as groups) to some product of the form   but since for all it evidently […]

    Pingback by Boolean Rings (Pt. I) « Abstract Nonsense | July 14, 2011 | Reply

  7. […] This follows trivially from the structure theorem. That said, there is a more simpleminded proof. Namely, by Cauchy’s Theorem there exists […]

    Pingback by A Classification of Integers n for Which the Only Groups of Order n are Cyclic (Pt. I) « Abstract Nonsense | September 13, 2011 | Reply

  8. […] the finitely generated abelian groups into their cyclic decomposition as is guaranteed by the structure theorem. Probably the most useful aspect of this post is that it will, if you keep your bookkeeping […]

    Pingback by Homomorphisms Between Finitely Generated Abelian Groups (Pt. I) « Abstract Nonsense | November 14, 2011 | Reply

  9. […] and spit this out. No. But, theoretically this does answer the pertinent question. Indeed, the structure theorem tells us that every finitely generated abelian group fits into the hypothesis criteria of this […]

    Pingback by Homomorphisms Between Finitely Generated Abelian Groups (Pt. II) « Abstract Nonsense | November 14, 2011 | Reply

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: