$G$ is an abelian group of order 32, how many $G$ are there such that every element $g\in G$, $g+g+g+g=e$? $G$ is an abelian group of order 32, how many $G$ are there such that every element $g\in G$, $g+g+g+g=e$?
According to the classification of abelian groups, we have 7 kinds of abelian groups of order $32$.
For instance, we have $Z_2 \oplus Z_2 \oplus Z_2 \oplus Z_2 \oplus Z_2$, which satisfies the requirement since every element(except zero) is of order $2$. One methond that I can think of is just examine every type individually. However, that is slow. Is there any other faster methods?
 A: If $(G,+)$ is a finite abelian group such that $4g=0\;$for all $g \in G$, each element must have order $1,2$, or $4$. It follows that $G$ is isomorphic to a direct sum of finitely many cyclic groups, where each summand is isomorphic to $Z_2$ or $Z_4$. 

But the order of a direct sum of finitely many groups is the product of the orders of the summands. Hence, if there are$\;a\;$summands isomorphic to $Z_2$, and $b$ summands isomorphic to $Z_4$, the order of the direct sum is $2^a4^b$, where$\;a,b\;$are nonnegative integers, at least one of which is positive.

But it's given that the order of$\;G$ is $32$, hence $2^a4^b = 32 = 2^5$.
\begin{align*}
\text{Then}\;\;&2^a4^b = 2^5\\[4pt]
\iff\;&2^a(2^2)^b=2^5\\[4pt]
\iff\;&2^a(2^{2b})=2^5\\[4pt]
\iff\;&2^{a+2b}=2^5\\[4pt]
\iff\;&a+2b=5\\[4pt]
\iff\;&(a,b) \in \{(5,0),(3,1),(1,2)\}\\[4pt]
\end{align*}
Hence there are $3$ qualifying isomorphism types.
A: You know the classification of finitely generated abelian group; in one of its formulations it says that a group $G$ of order $n$ is uniquely a direct sum of cyclic groups $G_1\oplus G_2\oplus\dots\oplus G_k$ in such a way that
$$
n_1\mid n_2\mid \dots \mid n_k
$$
where $n_i=|G_i|>1$ and, of course, $|G|=n_1n_2\dotsm n_k$. If the exponent of the group has to be at most $4$, we get that $n_k=4$ or $n_k=2$. Can you finish?
