Find the number of homomorphisms from $S_7$ to $A_8$
The kernel of a homomorphism $\phi: S_7 \to A_8$ is a normal subgroup of $S_7$.
I'd like to find all the homomorphisms by classifying the kernels of any such homomorphism.
$S_7$ has three normal subgroups: the whole group $S_7$, the trivial subgroup, and $A_7$.
This is the point where I'm not sure how to determine the exact number of all homomorphisms by kernels classification.
Homomorphisms with $\text{Ker}(\phi) = S_7$: If the kernel is the whole group $S_7$, then the homomorphism $\phi: S_7 \to A_8$ will be the trivial homomorphism, where every element of $S_7$ is mapped to the identity element in $A_8$. There is only one such homomorphism.
Homomorphisms with $\text{Ker}(\phi) = {e}$: If the kernel is the trivial subgroup ${e}$, this means the homomorphism is injective, and $S_7$ can be embedded into $A_8$. However, this is not possible since the alternating group $A_8$ has even permutations, while $S_7$ has odd permutations, and there is no subgroup of $A_8$ isomorphic to $S_7$. Therefore, there are no homomorphisms with $\text{Ker}(\phi) = {e}$.
Homomorphisms with $\text{Ker}(\phi) = A_7$ - this is the point where I'm stuck.
By the first isomorphism theorem for groups, the image of any homomorphism from $S_7$ to $A_8$ with kernel $A_7$ must have order $2$. Every homomorphism of this type will map odd elements of $S_7$ to some fixed element of order $2$ and even elements of $S_7$ to the identity of $A_8$.
As such, it remains to count how many elements of order $2$ there are in $A_8$. The only such elements possible are products of even amounts of disjoint transpositions. There are two possibilities: such an element could be a product of two disjoint transpositions or four disjoint transpositions.
If this element is a product of two disjoint transpositions, then there are $8\cdot 7\cdot 6\cdot 5$ choices for the numbers in the cycles. Because the cycles are disjoint, recognizing that $(x y)(a b) = (ab)(xy)$ and $(xy) = (yx)$, we must divide by $2\cdot 2^2$ to get $$\frac{8\cdot 7\cdot 6\cdot 5}{2\cdot 2^2} = 210$$ as the number of this type of element in $A_8$.
If this element is a product of four disjoint transpositions, then there are $8!$ choices for the numbers in the cycles. However, now we must divide by $4!\cdot 2^4$ by the same reasoning as above to avoid overcounting. This is because there are $4!$ ways to arrange the transpositions and $2$ ways to arrange the numbers inside each transposition. Thus we get $$\frac{8!}{2^4\cdot 4!} = 105$$ as the number of this type of element in $A_8$.
We conclude there are $315$ homomorphisms from $S_7$ to $A_8$ with kernel $A_7$.