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Not homework; reading a group theory book for fun.

This is what I understand so far:
If $\exists x \in G$ of order $9$ then $G \cong C_9$.
If not, then the order of $x$ has to divide $9$. So $\def\ord{\operatorname{ord}}\ord(x) = 1$ or $3$.
If $\ord(x) = 1$ then $x=e$.
So for $x$ of order 3, $\def\cyc#1{\left<#1\right>}\cyc x$ is a subgroup of order $3$.
Let $C$ be all $y \in G$ such that $xy=yx$. This is closed, has inverses, is associative, has identity, so is a subgroup.
The order of $C$ divides $9$ and is at least $3$ since $e, x, x^2$ commute with $x$. If $\ord(C)=9$ then choose $y \not\in\cyc x$.
$\cyc x \times \cyc y$ is $\{(x^a,y^b)\}$. The elements of $G$ are $e, x, x^2, y, y^2, xy, xy^2, x^2y, x^2y^2$. $x^ay^b \to (x^a,y^b)$ is an isomorphism. So $ G = C_3 \times C_3$
How do I continue this proof?


Edit: the next step in the book is to determine whether $\ord(C)=3$ for every choice of $x$. It then shows that $\exists x$ with order $9$.

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What do you mean continue it? It ends with the conclusion you would like. –  Tobias Kildetoft Nov 15 '13 at 8:34
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But you assumed that the group had no element of order $9$. So clearly all non-identity elements have order $3$. –  Tobias Kildetoft Nov 15 '13 at 8:43

2 Answers 2

You can even prove the general statement: Any group of order $p^2$ is isomorphic to either $\mathbb{Z}_{p^2}$ or $\mathbb{Z}_p \times \mathbb{Z}_p$.

Use Structure theorem for finite abelion groups.

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true, but the hard part here (though not terribly hard) is to show that any group of order $p^2$, $p$ prime, is abelian. –  Ittay Weiss Nov 15 '13 at 9:12

Your proof is just missing the treatment of the case that $C=\cyc x$, so that $|C|=3$ (your layout does not help the reader detect this omission).

If you know that any group of order $p^k$ with $p$ prime and $k>0$ has a non-trivial central element, then you can see that the omitted case cannot happen. Either some non-identity element of $\cyc x$ is central, but then $x$ is central (since every non-identity element of a subgroup of order $p$ generates it) and $C=G$, or there is a central element outside $\cyc x$, but then too $C$ is larger than just $\cyc x$.

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