Why is a simple group of order $60$ embedded in $A_6$?

Let $H$ be a simple group of order $60$. I am trying to see why it is embedded in $A_6$.

$H$ must have $6$ Sylow-$5$ subgroups and $H$ acting by conjugation on these subgroups gives an embedding of $H$ into $S_6$. But this is as far as i have gone.

Please give me only a hint.

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Identify $H$ with its embedded copy in $S_6$, and consider $H\cap A_6$. – Chris Eagle Oct 24 '11 at 15:18
@Chris: Done. If you want, make it an answer and i will vote it. – Manos Oct 24 '11 at 15:33

Identify $H$ with its embedded copy in $S_6$ and consider $H \cap A_6$.
Hint: there exists a canonical embedding of $S_5$ in $S_6$. That is, $S_5\leq S_6$. Why does this embedding imply that $A_5\leq A_6$?