Homomorphic image of $S_{8}$

Question

(1). Can $S_{8}$ be a homomorphic image of $S_{12}.$

(2). can $S_{8}$ be isomorphic to a subgroup of $S_{12}$

For 2 if we use the result that $S_{m}$ is isomorphic to a subgroup of $S_{n}$ iff $m \leq n$ then 2 is true. But what about 1.

Answer 1

You have the right idea for the second part. If you don't already know it, think about what the embedding $\Psi: S_m \hookrightarrow S_n$ when $m \leq n$ actually is; i.e. given an element $\pi \in S_m$, what element of $S_n$ is $\Psi(\pi)$? The map is quite "natural", so no need to overthink this.

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For the first part, suppose we have a surjective homomorphism $\phi: S_{12} \rightarrow S_8$. Applying the first isomorphism theorem, we have $S_{12}/\ker(\phi) \cong S_8$. The existence of this quotient group depends on the fact$^\dagger$ that the kernel of a group homomorphism $G \rightarrow G'$ is always a normal subgroup of $G$.

What would $|\ker(\phi)|$ need to be for the above isomorphism to hold? What are the possibilities for $\ker(\phi)$, i.e., the normal subgroups of $S_n$?

In greater generality, you can use this line of argument to classify every possible group homomorphism out of all of the symmetric groups. Be careful with $S_4$ though as it has a normal subgroup (isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$) that none of the other symmetric groups have.

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$^\dagger$As a side-note, the converse of this is also true: if $N$ is a normal subgroup of a group $G$, then there exists a homomorphism $\phi:G \rightarrow G'$ with $\ker(\phi) = N$, namely the natural projection $G \rightarrow G/N$ defined such that $g \mapsto gN$.