The number of homomorphisms from $S_5$ to $A_6$

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I used the fact that $A_5$ is the only real normal subgroup of $S_5$. Then we have a 2 types of homomorphisms and the trivial homomorphism.

  • homomorphisms with a kernel $\{e\}$

  • homomorphisms with a kernel $A_5$

In the first case the first isomorphism theorem implies that the image of the homomorphism is isomorphic to $S_5$, therefore the number of homomorphism of that type is the number of injections of $S_5$ in $A_6$.

In the second case because the index of $A_5$ in $S_5$ is 2, the quotient group is isomorphic to $C_2$. Therefore the image of the homomorphism is isomorphic to $C_2$.

That's as far as I got, I classified the homomorphisms but I don't know how to count them.

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There is no injective homomorphism from $S_5$ to $A_6$. Its image would have index $3$ in $A_6$. There is no such subgroup of $A_6$. If there were, there would be a homomorphism from $A_6$ onto a group of order $3$ or $6$, and there isn't.