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
Best Answer
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).