If $H$ is a subgroup of $S_5$ and $|H | = 8$ show that $H\cdot A_5 = S_5$

Suppose $H$ is a subgroup of $S_5$ and $|H|=8$. Show that $H\cdot A_5 = S_5$. Also determine $|H\cap A_5|.$

I think we can use the second isomorphism theorem to get the order of the intersection. I have no idea how to do the first part though. I believe $H$ must contain an odd permutation because it isn't a subgroup of $A_5$ by Lagrange's theorem. Is it true that: (some odd permutation)$\cdot A_5 = S_5?$

• Yes, it is true that $\{id, u\} \cdot A_5 = S_5$ for any odd permutation $u$: if $v\in S_5$ is even then $v\in A_5$, and if $v\in S_5$ is odd, then $u^{-1}v$ is even. – Catalin Zara Apr 17 '17 at 17:01

Order of $H$ does not divide order of $A_5$ so $H$ can not be contained in $A_5$, which means $H$ contains an odd permutation, so $H.A_5=S_5$.
Note that $H$ is a Sylow $2$-subgroup of $S_5$. All Sylow $2$-subgroups of $S_5$ are isomorphic to the dihedral group $$D_4=\langle (13),(1234)\rangle ,$$ because $(1234)(13)=(14)(23)=(13)(1432)=(13)(1234)^{-1}$. Using the odd permutation we see that $H\cdot A_5=S_5$.