# Find the number of homomorphisms from $S_7$ to $A_8$

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.

• I do not agree with the argument that there is no subgroup of $A_8$ isomorphic to $S_7$. This is true, but your argument is not correct (the image of an odd permutation under an embedding $S_n \to S_m$ does not have to be odd, e.g. you can embed $S_2 \hookrightarrow A_4$ by taking the non-trivial element to a product of disjoint transpositions). Commented Jul 4, 2023 at 23:07

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 $$A_7$$.