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Let $G$ be a non-abelian group of order $p^3$ for prime $p$. Show that $Inn(G)$ is abelian.

The center of $G$, $Z(G)$, is of order $p$ (can be seen in this question). I also know that $G / Z(G)=Inn(G) \Rightarrow \frac{p^3}{p}=p^2=|Inn(G)|$. How can I proceed and show $Inn(G)$ is abelian?

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    $\begingroup$ There are only two groups of order $p^2$: the cyclic group and $C_p\times C_p$. $\endgroup$ – Quang Hoang Mar 23 '16 at 2:22
  • $\begingroup$ @QuangHoang I'm not sure how that helps, sorry. $\endgroup$ – user3772119 Mar 23 '16 at 2:31
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    $\begingroup$ But did you just say that the order of $Inn(G)$ is $p^2$? $\endgroup$ – Quang Hoang Mar 23 '16 at 2:39
  • $\begingroup$ @QuangHoang oh yes, sorry. So if $Inn(G)$ is the cyclic group, then it is abelian? How do I show it's the cyclic group and not $C_p \times C_p$? $\endgroup$ – user3772119 Mar 23 '16 at 2:46
  • $\begingroup$ It's actually $C_p\times C_p$. $\endgroup$ – Quang Hoang Mar 23 '16 at 2:53
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We can prove that any group $G$ of order $p^2$ is abelian.

If $|Z(G)|=p^2$, then done. If $|Z(G)|=p$, then $G/Z(G)=p$ and so $G/Z(G)$ is cyclic. Suppose $hZ(G) (h\in G,h\notin Z(G))$ is the generator of $G/Z(G)$. Then for any $g_1,g_2\in G$, there is $g_1=h^nz_1, g_2=h^mz_2$, where $z_1,z_2\in Z(G)$. Thus $$ g_1g_2=h^nz_1h^mz_2=h^nh^mz_1z_2=h^mz_2h^nz_1=g_2g_1 $$

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  • $\begingroup$ I think you mean $h\not\in Z(G)$ while supposing $hZ(G)$ generates $G/Z(G)$, right? $\endgroup$ – pjs36 Mar 23 '16 at 4:04

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