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I was working on a exercise in Michael Artin's Algebra, which stated

A group $G$ of order 12 contains a conjugacy class of order $4$. Prove that the center of $G$ is trivial.

I started by noting that there must always be a conjugacy class of order $1$ (from the identity), and then we much have that the class equation can either be $1 + 1 + 4 + 6$ or $1 + 3 + 4 + 4$, since the order of the conjugacy classes must divide the order of the group. However, how would I eliminate the possibility of $1 + 1 + 4 + 6$ as a class equation, since this has nontrivial center.

EDIT: I just realized that my method was way off. Can anyone tell me a way to approach this problem?

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  • $\begingroup$ Why are conjugacy classes of order $2$ not possible? $\endgroup$ – Mark Bennet Nov 29 '14 at 20:52
  • $\begingroup$ Oops. I forgot about the other valid partitions of 12 - 4 - 1. $\endgroup$ – thkim1011 Nov 29 '14 at 20:54
  • $\begingroup$ One way of thinking about this would be to think of the group acting by conjugation on the class of size $4$ - the action is necessarily transitive. Your original group is nonabelian too (because it has a conjugacy class of size greater than $1$). Those are serious constraints. $\endgroup$ – Mark Bennet Nov 29 '14 at 21:05
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Let $G$ act on itself by conjugation. The orbit stabilizer theorem says $$ |G\cdot x|=\frac{|G|}{|\operatorname{Stab}(x)|}. $$ Here the orbit $G\cdot x$ is precisely the conjugacy class of $x$, denote it $C_G(x)$, and the stabilizer $\operatorname{Stab}(x)$ is precisely the centralizer of $x$, denote it $Z_G(x)$.

Let $x\in G$ have a conjugacy class of order $4$, then necessarily its centralizer $Z_G(x)$ has order $3$ by the above. But clearly $Z(G)\subseteq Z_G(x)$, so the center $Z(G)$ either has order $1$ or $3$. If $|Z(G)|=3$, then $x\in Z(G)$, but then it would have trivial conjugacy class as a central element, contrary to $|C_G(x)|=4$. So $Z(G)$ is trivial.

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  • $\begingroup$ Why does $Z(G) \subseteq Z_G(x)$ implies $Z(G)$ either has order $1$ or $3$? Why isn't it possible for it to have order $2$? $|Z(G)|$ must divide $|G|$ and indeed $2 \mid 12$. $\endgroup$ – Al Jebr Nov 19 '17 at 6:47
  • $\begingroup$ @AlJebr Also by Lagrange, $Z(G)$ is also a subgroup of the group $Z_G(x)$, so $|Z(G)|$ must also divide $3$. $\endgroup$ – Ben West Nov 19 '17 at 7:01

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