# 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.

## Facts used

1. Class equation of a group
2. Sylow subgroups exist
3. Sylow implies order-dominating: Any two Sylow subgroups are conjugate, and any $p$-subgroup is contained in a $p$-Sylow subgroup.
4. Cauchy's theorem for Abelian groups
5. 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 $Z(G)$:
1. 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 $p$.
2. 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 $p$.
2. Case that $p$ does not divide the order of $Z(G)$:
1. 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 $Z(G)$.
2. $C_G(g_i)$ 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 $p^r$.
3. $C_G(g_i)$ contains a subgroup of order $p^r$: This follows by fact (2).
4. 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).
5. 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 $C_G(gg_ig^{-1})$.