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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

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You are incorrect to claim that "$G$ must have an element $a$ of order $p$ by Lagrange's Theorem."

Lagrange's Theorem says that if $H$ is a subgroup of $G$, then $|H|$ divides $|G|$. In particular, by letting $H=\langle x\rangle$, it says that if $x$ is an element of $G$, then $|x|$ divides $|G|$.

Lagrange's Theorem does not say that if $p$ divides $|G|$, then there is an element of $G$ of order $p$.

You are very close, though.

Let $x$ be an element of $G$ other than the identity. By Lagrange's Theorem, you know that the order of $x$ must divide $|G|=p^2$, and since the only divisors of $p^2$ are $1$, $p$, and $p^2$, that means that $|x|=1$, $p$, or $p^2$. If it is $p^2$, you can proceed as you did above. If it is $p$, you can proceed as you did above. So why is it that you cannot have $|x|=1$?

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    Oh! That makes sense. Thank you so much.2011-10-13
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This follows from Cauchy's theorem. If $n$ is the group order and $p$ is a prime number dividing $n$, there is an element of order $p$ in $G$. In your case, $n=p^2$. Thus, there is an element of order $p$. The subgroup generated by this element is thus (cyclic) of order $p$.

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    Right, we are not that far yet. Thanks, though!2011-10-13
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Why would you consider the center of the group? Then you will find it commutative.