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I am stuck on the following exercise:

Let $G$ be a finite Abelian group that does not contain a subgroup isomorphic to $\mathbb Z_p \oplus \mathbb Z_p$ for any prime $p$. Prove that $G$ is cyclic.

This exercise comes before the chapter about fundamental theorem of finite Abelian groups and before the chapter about normal subgroups. Some of the chapters before the exercise are: cyclic groups, cosets and Lagrange's theorem, direct products.

So I assume there is a proof only using these tools (unless it is an accident and the exercise was meant to come later in the book).

Here is the idea I had:

Proof by contraposition: Assume $G$ is finite Abelian and not cyclic.

Then it contains two different subgroups of the same order (Although I'm not sure I can really deduce that since perhaps only the implication "cyclic implies contains exactly one subgroups per divisor of the order" holds)

Say they have order $n$. Let $p$ be a prime divisor of $n$. Now I want to deduce that each subgroup contains a copy of $\mathbb Z_p$ but I believe this might not be true.

Also, if we did have two disjoint copies of $\mathbb Z_p$ in $G$ then it's not clear to me how to get to $\mathbb Z_p \oplus \mathbb Z_p$ from there.

Please could someone tell me how to prove this?

Thomas Andrews
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  • If $H$ and $K$ are two subgroups that are distinct and of order $p$, then since G is abelian, $HK$ is a subgroup and it must be of order $p^{2}$ since $H \cap K$ must be identity or else $H =K$. But then $HK$ is an abelian group of order $p^{2}$. Since it cannot be the type prohibited in the problem, $HK$ must be the cyclic group of order $p^{2}$. But $HK$ contains both $H$ and $K$ and this is a contradiction. – Daniel Akech Thiong Feb 20 '16 at 02:41

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