Every Sylow subgroup intersects the center nontrivially or is contained in a centralizer

Statement
Suppose $$G$$ is a finite group and $$p$$ is a prime dividing the order of $$G$$. Then, every $$p$$-Sylow subgroup of $$G$$ satisfies at least one of these two conditions:


 * 1) It intersects the center of $$G$$ nontrivially.
 * 2) it is contained in the centralizer of a non-central element.

Further, if any one $$p$$-Sylow subgroup satisfies a particular condition, so do all the others.

Related facts

 * Sylow subgroups exist: In fact, one of the proofs of the existence of Sylow subgroups essentially uses this idea.
 * Sylow implies order-conjugate: Any two $$p$$-Sylow subgroups are conjugate.

Facts used

 * 1) uses::Class equation of a group
 * 2) uses::Sylow subgroups exist
 * 3) uses::Sylow implies order-dominating: Any two Sylow subgroups are conjugate, and any $$p$$-subgroup is contained in a $$p$$-Sylow subgroup.
 * 4) uses::Cauchy's theorem for Abelian groups
 * 5) uses::Central implies normal

Proof
Given: A finite group $$G$$ of order $$n = p^rm$$, where $$p$$ is prime, $$r$$ is a positive integer, and $$p$$ does not divide $$m$$.

To prove: Every $$p$$-Sylow subgroup of $$G$$ either intersects the center nontrivially, or is contained in the centralizer of a non-central element.

Proof: Consider the class equation of $$G$$ (fact (1)):

$$|G| = |Z(G)| + \sum_{i=1}^r |G:C_G(g_i)|$$

where $$c_1,c_2,\dots,c_r$$ are the conjugacy classes of non-central elements and $$g_i$$ is an element of $$c_i$$ for each $$i$$.

We consider two cases:


 * 1) Case that $$p$$ divides the order of $$Z(G)$$:
 * 2) There exists a normal subgroup of order $$p$$ in $$G$$: Since $$Z(G)$$ is Abelian, fact (4)  yields that it has a subgroup $$H$$ of order $$p$$. Since $$H$$ is in the center, $$H$$ is normal in $$G$$ (by fact (5)). Thus, $$H$$ is a normal subgroup of $$G$$ of order $$p$$.
 * 3) Suppose $$P$$ is any $$p$$-Sylow subgroup of $$G$$. By fact (3), the subgroup $$H$$ is contained in some conjugate of $$P$$. Since $$H$$ is normal, this forces $$H \le P$$. Thus, $$P$$ intersects the center nontrivially -- the intersection contains a subgroup of order $$p$$.
 * 4) Case that $$p$$ does not divide the order of $$Z(G)$$:
 * 5) There exists $$i$$ such that $$p$$ does not divide the index $$k$$ of $$C_G(g_i)$$ in $$G$$: Since $$p$$ divides the order of $$G$$, $$p$$ cannot divide the index of every $$C_G(g_i)$$, otherwise the class equation would yield that $$p$$ divides the order of $$Z(G)$$.
 * 6) $$C_G(g_i)$$ is a proper subgroup of $$G$$ whose order is a multiple of $$p^r$$: Since $$g_i$$ is non-central, $$C_G(g_i)$$ is proper in $$G$$. Further, since $$|G:C_G(g_i)| = k $$ is relatively prime to $$p$$, Lagrange's theorem (fact (3)) yields that the order of $$C_G(g_i)$$ is $$p^rm/k$$, which is a multiple of $$p^r$$.
 * 7) $$C_G(g_i)$$ contains a subgroup of order $$p^r$$: This follows by fact (2).
 * 8) Any $$p$$-Sylow subgroup of $$G$$ is of the form $$gPg^{-1}$$ for some $$P \le C_G(g_i)$$: This follows from the previous step, and fact (3).
 * 9) Any $$p$$-Sylow subgroup of $$G$$ is contained in the centralizer of a non-central element: This follows from the previous step; in fact, it is contained in $$C_G(gg_ig^{-1})$$.