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

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.

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.
• If this argument were correct, it would also show that there's no injective homomorphism from $S_5$ to $A_7$. But there is such a homomorphism: Send every even permutation of $\{1,2,3,4,5\}$ to itself, leaving $6$ and $7$ fixed, and send every odd permutation of $\{1,2,3,4,5\}$ to its product with the transposition $(67)$. Feb 7 '18 at 19:29