3
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ 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). $\endgroup$
    – hunter
    Commented Jul 4, 2023 at 23:07

1 Answer 1

2
$\begingroup$

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 $A_7$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .