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.

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*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:


*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,
 A: @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.
A: 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.
