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I'm working on a problem from Artin which asks to rule out as many class equations for a group of order 10 as I can. I'm are unsure about one:

1 + 1 + 2 + 2 + 2 + 2

My thought was that there are four conjugacy classes of order 2 each, so their corresponding centralizers will each have order five. Since they have prime order, we know the centralizers are all cyclic, so that implies they're the same. But what kind of group would have four conjugacy classes with the same centralizer?

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  • $\begingroup$ @DonAntonio Well that just means they are in the center, correct? And since there are 2, which divides 10, that's okay as far as I know. $\endgroup$ – Phdetermined Sep 26 '13 at 0:13
  • $\begingroup$ Of course. I misunderstood. $\endgroup$ – DonAntonio Sep 26 '13 at 0:17
  • $\begingroup$ Can you use Sylow theorems and etc.? $\endgroup$ – DonAntonio Sep 26 '13 at 0:19
  • $\begingroup$ @DonAntonio These exercises come before the Sylow theorems are introduced, so I figured there was a way to figure it out without them. Up until now, I have the class equation and counting formula which are what I was working with. $\endgroup$ – Phdetermined Sep 26 '13 at 0:20
  • $\begingroup$ Ayman just posted an excellent answer, which assumes you already know, or can prove, that for any group, its quotient over its center cannot be cyclic non-trivial $\endgroup$ – DonAntonio Sep 26 '13 at 0:23
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If $\left|Z(G)\right| = 2$, then $|G/Z(G)| = 5$. Thus $G/Z(G)$ is cyclic. Hence $G$ is abelian by this theorem. This contradicts with $|Z(G)| = 2$. It follows that no group has the class equation you have.

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