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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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    $\begingroup$ 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 $G^{p^2}=\{e\}$. $\endgroup$ Oct 13 '11 at 21:17
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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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  • $\begingroup$ When you say x is an element of G, you're also saying that x is a subgroup of G, right? $\endgroup$
    – Lily
    Oct 13 '11 at 21:21
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    $\begingroup$ @Lily: No, I'm saying $x$ is an element of $G$. Why would I imply that it is a subgroup? Elements are not usually subgroups. Of course, given an element you can always form "the subgroup it generates", but the element and the subgroup are different animals (though of course intimately connected to one another). $\endgroup$ Oct 13 '11 at 21:22
  • $\begingroup$ Oh, I thought LaGrange's Theorem said that the order of the subgroup divides the order of the group. $\endgroup$
    – Lily
    Oct 13 '11 at 21:25
  • $\begingroup$ @Lily: Yes, that's what it says, and that's what I quoted. That is not what you tried to use it for, though! You tried to invoke it to go from "this number divides the order of the group" to "there is an element whose order is this number." Lagrange's Theorem goes the other way, from "there is an element whose order is this number" to "this number divides the order of the group." $\endgroup$ Oct 13 '11 at 21:27
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    $\begingroup$ @Lily: Yes: the only element of a group that can have order $1$ is the identity. But we started the whole chain of reasoning saying "Let $x$ be an element of $G$ other than the identity". So $x\neq e$, which means $|x|\neq 1$. $\endgroup$ Oct 13 '11 at 21:39
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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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    $\begingroup$ I would guess that this problem is being posed before proving Cauchy's Theorem... $\endgroup$ Oct 13 '11 at 21:05
  • $\begingroup$ Right, we are not that far yet. Thanks, though! $\endgroup$
    – Lily
    Oct 13 '11 at 21:23
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Why would you consider the center of the group? Then you will find it commutative.

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