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Let $G$ be a group of order $p^2$. Use the class equation to prove that $G$ is abelian.

The conjugacy class equation, at least how I remember it, is $$ |G| = |Z(G)|+\sum_{x\in I \backslash Z(g)} \frac{|C|}{C_G(x)} $$ where $|Z(G)|$ is the center of $G$, and $|C_G(x)|$ is the central subgroup of $G$.

I am tempted to start with using Lagrange's theorem, which gives possible orders $o(x,y)$ of $1,p,p^2$.

Now, to prove that $G$ is abelian, perhaps I need to prove that $|C_G(x)|=G$? ($C_G(x)$ being the central subgroup in which every element inside of it commutes)

Cookie
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1 Answers1

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In order to prove that $G$ is abelian you need to prove that $G=Z(G)$. Now possible orders for $Z(G)$ are $1,p,p^{2}$. Since $|Z(G)|$ must be divisible by $p$ so $|Z(G)|\neq 1$. If $|Z(G)|=p$ then let $a\in G$ such that $a$ is not in $Z(G)$. Then $Z(G)\subset C(a)$ and $a\in C(a)\ \Rightarrow$ $|C(a)|>p$ but $|C(a)|$ divides $p^{2}$ and hence $|C(a)|=p^{2}$ and therefore $a\in Z(G)$, a contradiction. Hence $|Z(G)|=p^{2}$ and so $G$ is abelian.

Cookie
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wanderer
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  • I liked your explanation which utilizes the Lagrange's theorem beautifully. But I really wanted to know how the class equation can be applied to prove the same result. – Cookie Mar 19 '14 at 06:32
  • the fact that we are discarding the possibiity of $|Z(G)|\neq 1$ is guranteed by the class equation. – wanderer Mar 19 '14 at 06:34