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Let $G$ be a group of order $p^2$, where $p$ is prime. Show that $G$ must have a subgroup order of order $p$.

What I have so far:

$$G^{p^2} =e .$$

If $G$ has an element $g$ of order $p^2$, then $g^p$ is of order $p$. $\langle g^p\rangle$ is a subgroup of order $p$.

$G$ must have an element $a$ of order $p$ by Lagrange's Theorem. $\langle a\rangle$ is a subgroup of order $p$.

Is this sufficient? Or am I missing some details?

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    By the way: $G^{p^2}=e$ is somewhat abusive of notation. The left hand side is the subgroup of $G$ generated by all elements of the form $g^{p^2}$, with $g\in G$; the right hand side is a single element. It would be better to write $G^{p^2}=\{e\}$.2011-10-13

3 Answers 3