# Showing that 2-Sylow subgroups intersect trivially [duplicate]

I'm looking over a proof for something and I can't understand the last bit of a certain part (counting argument).

$G$ is a group of order 80 and $n_2 = 1$ or $5$ (number of 2-Sylow subgroups). We assume that it's not one so it is five. Then let $P_1 \neq P_2$ be 2-Sylow subgroups with $D$ being the intersection of these. So, we know that $P_1, P_2 \in C_G(D)$ but if the centralizer has two 2-Sylow subgroups it must have at least 1+2=3 2-Sylow Subgroups. So, $|C_G(D)| \ge 16(3) = 48 \implies C_G(D) = G$. I am fine with all of this, just the final bit... this is a contradiction. Am I missing some obvious reason that $C_G(D) = G$ is a contradiction?

Then we can conclude that the subgroups intersect trivially, which is what I want.

EDIT: The objective of the proof is simply to show that either the 2-Sylow or 5-Sylow subgroup must be normal. There is no mention of the group being abelian or non-abelian, so this is unknown. Before this part we assume that $n_5 = 16$ giving 64 elements of order 5.

Thanks!

## marked as duplicate by Dietrich Burde, Community♦Jan 4 '18 at 19:35

• Indeed, if $G$ is abelian, then $C_G(D)=G$ is not a contradiction. – Dietrich Burde Jan 4 '18 at 19:08
• The objective of the proof is simply to show that either the 2-Sylow or 5-Sylow subgroup must be normal. There is no mention of the group being abelian or non-abelian, so this is unknown. – Helen Byrne Jan 4 '18 at 19:11
• Then we cannot get a contradiction without a further argument, because $G$ could be abelian, too. But we can easily see that there is a normal $2$-Sylow group, see below. – Dietrich Burde Jan 4 '18 at 19:14
• How did you get to $P_1,P_2\subseteq C_G(D)$? We don't know that $P_1$ and $P_2$ are abelian, so there is no reason to assume that they would centralize $D$ – Jyrki Lahtonen Jan 4 '18 at 19:38
• I think this proof provided by the lecturer was totally incorrect upon reflection. The argument that there are 64 elements of order 5 and only room for one 2 Sylow as per Dietrich is far more straightforward. – Helen Byrne Jan 5 '18 at 22:40

I would propose a different counting argument. Since the $5$-Sylow groups have order $5$, they are cyclic. So any two elements of it intersect trivially. Therefore $G$ has at most $80 – 4 ⋅ 16 = 16$ elements whose orders are powers of $2$, the exact number of elements in a Sylow $2$-subgroup. This means that there is a unique Sylow 2-subgroup of $G$, which then is a normal subgroup.