Isomorphisms of finite abelian groups 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,
kasp9201.
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
 A: 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).
A: 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.
