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- Problem about solvable groups 1 answer
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.