Let $G \cong A_6$ and suppose that $G$ acts as a transitive permutation group on a set $\Omega$ of size $45$. I want to prove that every involution fixes five points. Any ideas how this could be done?
We have $45 = |G : G_{\alpha}|$, hence $|G_{\alpha}| = 8$ is a Sylow $2$-subgroup. So this action is equivalent to the action on some Sylow $2$-subgroup of $A_6$. For example take $S = \langle (1 2 3 4)(5 6), (1 3)(5 6) \rangle \cong D_8$ from here. As $Sxy = Sx \Leftrightarrow y \in S^x$ the statement is equivalent to the statement that every involution $y$ lies in $S^{x_1} \cap S^{x_2} \cap S^{x_3}\cap S^{x_4} \cap S^{x_5}$, where $x_1, \ldots, x_5$ represent different cosets, and not more. But if this might be the right approach, how to show that the double transposition $y$ (the involutions in $A_6$ are precisely the double transpositions) lies in precisely five different conjugates of $S$?
The elements of $S$ are $$ 1, (12)(34), (1 3)(2 4), (1 4)(2 3) , (1 2 3 4)(5 6), (1 4 3 2)(5 6), (1 3)(5 6), (24)(56). $$ So you see that you can parametrize all these Sylow $2$-subgroup $S$ in terms of the unordered pairs that play the role of $5, 6$ (of which there are $15$), and in terms of the number subgroups of order $4$ on the remaining numbers $1, 2, 3, 4$ (that is, $3$).
So call the $S$ above $S(56, 1234)$. For instance $$ S(34,1526) = \{1, (15)(26), (12)(56), (16)(25), (1526)(34), (1625)(34), (12)(34), (56)(34) \} $$
Now it is not difficult to see that $(12)(34)$, say, is contained exactly in $$ S(56, 1234), S(56, 1342), S(56,1423), S(34,1526), S(12,3546). $$