Given $|G|=p^2$ then how can you deduce $G\cong C_{p^2}$ or $G \cong C_p \times C_p$
I have shown that G is abelian, not sure what to do next
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user35603
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The Problem
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2Look for an element of order $p^2$ (equivalently: decide if $G$ is cyclic or not). – David Wheeler Mar 01 '15 at 20:36
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Use the fundamental theorem of finitely generated abelian groups, which states that every such group is a product of cyclic groups.
If you cannot use this, note that every element has either order $p$ or order $p^2$. If an element has order $p^2$, then we have a cyclic group of order $p^2$. Otherwise every element has order $p$. Choose two elements neither of which is a power of the other; then the subgroups generated by these elements are normal, intersect trivially, and generate the whole group (because the subgroup generated by the two elements must have order $p^2$), hence we have a direct product of two cyclic groups of order $p$.
Matt Samuel
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Could you explain how more precisely how to show the second part (the case where $o(g)=p$) I'm not sure to understand why the two chosen elements intersect trivially, and generate the whole group. – Sylvester Stallone Jun 16 '17 at 13:48
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@Syl The subgroups of order $p$ are cyclic and every nonidentity element is a generator. Thus if two subgroups of order $p$ intersect nontrivially then they are the same. – Matt Samuel Jun 16 '17 at 14:19