Determining the center of the p-Sylow subgroup of $S_p $ My Algebra book says without proof that the center of the p-Sylow subgroup (we will call it P) of $S_p$ is the subgroup itself. 
Now I can understand that P will be a subset of its center as it is of prime order, and hence cyclic. However, I can't understand why no other element outside of P can be part of its center. Any help understanding this would be great. 
 A: Like anon I assume that the question is about the centralizer of $P$ (with that typo/mistranslation the question makes IMHO more sense).
The highest power of $p$ that divides $p!$ is $p^1$. Therefore the Sylow subgroups $P$ are all cyclic of order $p$. A permutation of prime order $p$ must be a product of disjoint $p$-cycles (ignoring fixed points). In the case of $S_p$ there is room for only one disjoint $p$-cycle, so we know that $P$ is generated by a $p$-cycle, call it $\sigma$. Because $P=\langle\sigma\rangle$ the centralizer of $P$ equals the centralizer of $\sigma$, $C_{S_p}(\sigma)$.
Let's take a detour via conjugacy classes. We know from basic properties of symmetric groups that the conjugates of $\sigma$ are exactly all the $p$-cycles in $S_p$. How many are there? A $p$-cycle in $S_p$ will move all the numbers $1,2,3,\ldots,p$. Recall that in cycle notation we can choose the starting point of the cycle any way we want without changing the cycle, but no other changes are possible. So we can write all $p$-cycles $\alpha$ of $S_p$ uniquely in such a way that they begin with a $1$. We then have $p-1$ choices for the next number $=\alpha(1)$, $p-2$ choices for the next $=\alpha(\alpha(1))$ et cetera. Therefore there are exactly $(p-1)!$ different $p$-cycles in $S_p$. In other words, the conjugacy class $[\sigma]$ of $\sigma$ has size $(p-1)!$.
But the orbit-stabilizer theorem (often 
introduced at about the same time as the class equation) tells us that
$$
(p-1)!=|[\sigma]|=\frac{|S_p|}{|C_{S_p}(\sigma)|}=\frac{p!}{|C_{S_p}(\sigma)|}.
$$
This implies that the centralizer $C_{S_p}(\sigma)$ has order $p$. Clearly $P\subseteq C_{S_p}(\sigma)$, so this means that we must have equality
$$
P=\langle\sigma\rangle=C_{S_p}(\sigma)=C_{S_p}(P).
$$
