A finite abelian group in which each element has order 1 or $p$, $p$ prime, is an elementary abelian group

Question

Prove: a finite abelian group $G$ in which each element has order 1 or $p$, $p$ prime, is an elementary abelian group.

Definition: Let $p$ be a prime number and $n\ge 1$. Then $E_{p^n}:=C_p \times C_p \times\cdots\times C_p$ is an elementary abelian group of order $p^n$.

My attempt:

Let $n=|G|$. Consider $g\ne e \in G$, then $\langle g \rangle \cong C_p \trianglelefteq G$ since $G$ is given to be abelian. Choose $h\ne g\ne e\in G$, and let $x\in \langle g\rangle \cap \langle h\rangle$. Element $x$ has order $p$, thus $\langle x \rangle = \langle g\rangle \cap \langle h\rangle \cong C_p$.

I don't see how I can conclude that $\langle g\rangle \cap \langle h\rangle$ is trivial. Then I'd also have to show that $\prod_{g\in G} \langle g\rangle = G$, to find that $G\cong C_p \times \cdots\times C_p$.

Any ideas?

Thanks.

Answer 1

Choose $h\notin \langle g\rangle$. Then, by Lagrange's theorem, $\langle g\rangle \cap \langle h\rangle$ is trivial.

If $G$ is finite, then the process of repeatedly choosing elements not in the product that you have at that stage must stop and will give you the direct product you require.

Answer 2

Clearly $|G|=p^n$ for some $n$. Take $H\leq G$ with $|H|=p^{n-1}$. Observe $G/H$ is cyclic and let $x\in G$ such that $G/H=\langle xH\rangle$. Consider the map $\phi:H\times G/H\to G$ given by $$(h,x^kH)\mapsto hx^k.$$ Note that $\phi$ is well defined, since $$x^kH=x^mH\implies x^{k-m}\in H\implies x^{k-m}\in H\cap\langle x\rangle=\{1\} \implies x^{k-m}=1\implies x^k=x^m.$$

$\phi$ is clearly surjective thus bijective by the finitude of $G$. Moreover it's a group homomorphism: $$\phi(h,x^kH)\phi(g,x^mH)=(hx^k)(gx^m)=(hg)x^{k+m}=\phi(hg,x^{k+m}H)=\phi((h,x^kH)(g,x^mH)).$$ Hence $G\cong H\times G/H\cong H\times C_p$. Repeat this process with $H$ and so forth until achieving the desired result.