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So I've been working through Beachy/Blairs Abstract Algebra book, and on the last few sections I seem to get continually hung up on questions dealing with centralizers.

  1. The last one I encountered was, "Show that if $ n \ge3 $, then then center of $ S_n $ is trivial."

    I was able to do this by contradiction and by exploiting something I knew about permutation groups, but for general groups I have an absolutely horrid time working them:

  2. Let G be a group and let $a \in G $. Show $C(a)$ is a subgroup of G and show $\langle a \rangle \subseteq C(a) $.

What are some good things to think about when I go to solve problems like these?

Thanks,

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    @MichaelJoyce: It was Chris's advice; I was just commenting that it was generalisable. While I agree that it applies to all maths, I don't think it's advice that anyone beyond undergrad level needs. Also there is a big difference at research level, as indicated by what you said about 'immense amounts of time understanding definitions $\ldots$'. Anyway, this is probably getting too discussiony for comments.2012-04-25

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@Andrew W., observe that for a group $G$ the center $Z(G) = \bigcap_{g \in G} C_G(g)$. Hence if you start calculating some of the centralizers, you will get a hint of which elements are in the center and which elements are not.

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It is sometimes helpful to understand the centralizer of a subgroup $H\leq G$ as the kernel of an action, since this tells you that it is always a subgroup of $G$, and sometimes a normal subgroup of $G$.

First consider the normalizer of $H$ in $G$:

$N_G(H) = \{ g \in G \mid gHg^{-1} \subseteq H \}.$

Since $H$ is stable under conjugation by elements of $N_G(H)$, you can define the conjugation action of $N_G(H)$ on $H$, that is, $\phi_g(h) = g h g^{-1}$. The map $g \mapsto \phi_g$ is a homomorphism of $N_G(H)$ into $\mathrm{Aut}(H)$, and the kernel of $\phi$ is

$C_{N_G(H)}(H) = \{g\in N_G(H) \mid ghg^{-1}=h, \; \forall h\in H\}.$

Now note that $C_{N_G(H)}(H) = C_G(H)$, the centralizer of $H$ in $G$.

This reveals that $C_G(H)$, being the kernel of a hom, is a normal subgroup of $N_G(H)$, so it is a subgroup of $G$. If $H$ happens to be a normal subgroup of $G$, then $C_G(H)\trianglelefteq N_G(H)=G$, so $C_G(H)$ is normal in $G$ in this case.