I'm studying Hungerford's "Abstract Algebra - An Introduction". In its chapter 9.2 Hungerford gives an example of a characterization up to isomorphism of all finite groups of order 36. For this he uses the Fundamental Theorem of Finite Abelian Groups. The theorem states:

"Every finite abelian group $G$ is the direct sum of cyclic groups, each of prime power order",

where for a cyclic group with order $p^n$, we have that $p \vert m$, and $m$ is the order of $G$. The example I refer to is as follows:

"The number 36 can be written as a product of prime powers in just four ways: $36 = 2\cdot 2 \cdot 3 \cdot 3 = 2 \cdot 2 \cdot 3^2 = 2^2 \cdot 3 \cdot 3 = 2^2 \cdot 3^2$. Consequently, every abelian subgroup of order 36 must be isomorphic to one of the following groups:

$\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_3$,

$\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_9$,

$\mathbb{Z}_4 \times \mathbb{Z}_3 \times \mathbb{Z}_3$,

$\mathbb{Z}_4 \times \mathbb{Z}_9$

These are easily shown to not be isomorphic to each other by examining their elements. Furthermore, $\mathbb{Z}_{36}$ is isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_9$."

All this I understand. But Hungerford argues that this is a complete characterization of all finite abelian groups of order 36 up to isomorphism. This, I don't understand. How do we guarantee that no other isomorphisms exist? Thank you, and best regards,


Edit: Here is a clarification of my question. I understand that a finite abelian group of order 36 is indeed isomorphic to the four direct products that I have listed. How do we guarantee that a finite abelian group of order 36 is not isomorphic to more than just these four? A finite abelian group $G$ can be written as the direct product of p-groups $G(p_1) \times G(p_2) \times ... \times G(p_n)$, where if $\vert G \vert = m$, then $p_i \vert m$ for all $i$. Each of these groups $G(p_i)$ can then be written as direct products of cyclic subgroups $(k)$, where $\vert k \vert = p_i^c$, if $k \in G(p_i)$.

It seems to me that Hungerford's argument is that since a finite abelian group $G$ is isomorphic to the direct product of such cyclic subgroups, then it is only isomorphic to such cyclic subgroups. That is, if $G$ was to be isomorphic to any other group, then this group would be isomorphic to one of $G$'s direct products of cyclic subgroups. Is this what Hungerford (and lhf) builds his argument on? If so, why is this true?

Thank you again, and I apologize for any inconvenience

  • 2
    $\begingroup$ The key fact is "The number 36 can be written as a product of prime powers in just four ways". $\endgroup$
    – lhf
    Apr 9, 2018 at 13:36
  • 3
    $\begingroup$ This is a classification of abelian groups of order $36$, not of all groups of order $36$. There are of course several isomorphism classes of nonabelian groups of order $36$. $\endgroup$
    – Derek Holt
    Apr 9, 2018 at 13:37
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    $\begingroup$ What you call the Fund. Theo. of Finite Abelian Groups is not even a theorem, as it is false. For example, the finite abelian group $\;\Bbb Z_4=\Bbb Z/4\Bbb Z\;$ is not a direct sum of cyclic group of order a prime . Very important: to read, understand an quote correctly theorems in mathematics. $\endgroup$
    – DonAntonio
    Apr 9, 2018 at 13:39
  • 1
    $\begingroup$ You are still talking about all groups of order $36$ when you mean all abelian groups. It is difficult to tell exactly what it is that you do not understand. $\endgroup$
    – Derek Holt
    Apr 9, 2018 at 13:50
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    $\begingroup$ So you appear to have written down a correct proof that there are exactly four isomorphism classes of abelian groups of order $36$ and then written "all this I understand", but then you wrote: "This I don't understand"! Could you explain more clearly exactly what it is that you do not understand? $\endgroup$
    – Derek Holt
    Apr 9, 2018 at 14:05

2 Answers 2


There are two characterization of finite abelian groups by invariants.

The simplest to manage is that a finite abelian group $G$ of order $n$ can be uniquely written (up to isomorphism) as a direct product of cyclic groups $G=G_1\times G_2\times\dots\times G_k$, with $|G_i|=n_i>1$ and $$ n_1\mid n_2\mid \dots \mid n_k $$ (of course $n_1n_2\dots n_k=n$). This is called the invariant factor decomposition.

For $n=36$ we can have \begin{align} &2,18\\ &3,12\\ &6,6\\ &36 \end{align} that is, just four choices. If $C_m$ denotes the cyclic group of order $m$, we can also get the other decomposition (primary decomposition): \begin{align} &C_2\times C_{18}\cong C_2\times C_2\times C_9\\ &C_3\times C_{12}\cong C_3\times C_3\times C_4\\ &C_6\times C_6\cong C_2\times C_2\times C_3\times C_3\\ &C_{36}\cong C_4\times C_9 \end{align}

The fact that these two classifications are complete is a quite deep theorem proved by Kronecker (1870).

  • $\begingroup$ Hi egreg, thank you very much for having taken your time. I've thought about both answers given for a couple of hours now, and they both add nice perspectives on what I wanted. Unfortunately I have to choose an answer, so I picked the one given by D_S by the fact that I figured out my question when reading his answer. But again, thank you very much and have a nice day $\endgroup$
    – kasp9201
    Apr 9, 2018 at 16:42
  • $\begingroup$ Accidently I find your answer helpful once again in MSE rather than TSE. $\endgroup$
    – C.F.G
    Sep 8, 2022 at 6:16

Let $G$ be a finite abelian group of order $36$. The fundamental theorem tells us that there exist cyclic subgroups $H_1, ... , H_t$ of prime power order such that $G$ is equal to the (internal) direct sum of $H_1, ... , H_t$. Thus $G = H_1 + \cdots + H_t$ and $H_i \cap (\sum\limits_{j \neq i} H_j) = 0$. The internal and external direct sums are isomorphic. To be more specific, the map sending $(h_1, ... , h_t)$ to $h_1 + \cdots + h_t$ defines a group isomorphism

$$H_1 \times \cdots \times H_t \rightarrow G$$

Each $H_i$ is cyclic of prime power order, say $H_i$ has $p_i^{e_i}$ elements for some prime number $p_i$ and positive integer $e_i$. Then

$$36 = |G| = |H_1| \cdots |H_t| = p_1^{e_1} \cdots p_t^{e_t}$$

By the fundamental theorem of arithmetic, there are only finitely many possibilities of prime numbers $p_i$ and exponents $e_i$ such that the $p_i^{e_i}$ multiply to $36$. Without loss of generality, let us assume that $p_1 \leq \cdots \leq p_t$. Then you can check yourself that the only possibilities for the number $t$, the prime numbers $p_i$, and the exponents $e_i$ are:

$$t= 4, p_1 = p_2 = 2, p_3 = p_4 = 3, e_1 = e_2 = e_3 = e_4 = 1$$

$$t = 3, p_1 = p_2 = 2, p_3 = 3, e_1 = e_2 = 1, e_3 = 2$$

$$t = 3, p_1 = 2, p_2 = p_3 = 3, e_1 = 2, e_2 = e_3 = 1$$

$$t = 2, p_1 = 2, p_2 = 3, e_1 = 2, e_2 = 2$$

Let us suppose we have the third possibility. Then $H_1 \times H_2 \times H_3 \rightarrow G, (h_1, h_2,h_3) \mapsto h_1 + h_2 + h_3$ is a group isomorphism, where $H_1$ is cyclic of order $4$, and $H_2$ and $H_3$ are cyclic of order $3$. All cyclic groups of a given order are isomorphic, so there exist group isomorphisms $\mathbb{Z}_4 \rightarrow H_1, \mathbb{Z}_3 \rightarrow H_2, \mathbb{Z}_3 \rightarrow H_3$. Combining these isomorphisms gives you an isomorphism

$$\mathbb{Z}_4 \times \mathbb{Z}_3 \times \mathbb{Z}_3 \rightarrow H_1 \times H_2 \times H_3 \rightarrow G$$

so $G$ is isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_3 \times \mathbb{Z}_3$. The other possibilities for $t, p_i,e_i$ would show that there exists an isomorphism of $G$ with the other groups.

  • $\begingroup$ Hi @D_S. As you can see, I have chosen your answer as an answer to my question. It gave me the intuition I was looking for, so thank you very much! have a nice day $\endgroup$
    – kasp9201
    Apr 9, 2018 at 16:49

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