Show that a group of order 12 cannot have nine elements of order 2.

Question

This is what I have so far. Assume the contrary, then by Sylow's Third Theorem, the number of Sylow 3-subgroups is either 1 or 3. If it were 1 then there are only 7 elements of order 2. Thus there are 3 Sylow 2-subgroups. I am trying to wrap this up but I am getting confused. Is it the case that each Sylow 2-subgroup has the identity and 7 elements of order 2?

Answer 1

Such a group would have two elements of order $3$. Let $h$ be one of them. The centraliser of $h$ must have order $6$ or $12$. In any case this centraliser contains an element of order $2$ whose product with $h$ has order $6$; that's impossible.