Centers of symmetric groups

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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This question is just juggling definitions. Just write down the definitions of $C(a)$ and of $\langle a \rangle$, write down what "is a subgroup of $G$" means and what "$\subseteq$" means, and use the definitions to prove the things you have to prove one by one. –  Chris Eagle Apr 22 '12 at 20:48
Great! Thanks. It looks really obvious now. –  Andrew W. Apr 22 '12 at 21:01
You can apply that advice to most of undergraduate maths. If you have no idea how to approach a problem, start by writing down the definitions of everything in the question, and most of the time that will help a lot. –  Tara B Apr 23 '12 at 8:20
@AndrewW.: How about you write your solution as an answer now? It's not something I've seen done a lot on this site, but it's nice to have questions not hanging around as unanswered unnecessarily, and also then people can let you know if your proof is correct. –  Tara B Apr 23 '12 at 8:22
@TaraB: I really like your advice, but I would say it applies to all mathematics, even the most advanced! Of course, insights beyond merely unpacking definitions are the signs of exceptional craftmanship, but geniuses often spend immense amounts of time understanding definitions and logical implications before they achieve their breakthroughs (or so I've read). –  Michael Joyce Apr 24 '12 at 20:46

@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.

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