How to show isomorphism of permutation groups?

Question

Let $M = \{m_1,. . . , m_k\} ⊂ \{1, 2, 3,. . . , n\}$ and $H = \{\sigma ∈ S_n | \forall m \in M, \; \sigma(m) \in M\}$.

Show that $H \subset S_n$ is a subgroup.

I tried the following: $m \in M$ and $\sigma \in H$ implies that $\sigma(m) \in M$. If we take $\pi \in H$, then $\pi \sigma = \pi(\sigma(m)) \in M$ which in turn implies that $\pi\sigma \in H$(here the product is defined as a composition).

Update: I have shown that $H$ is nonempty, finite and closed under the operation of multiplication, hence I have shown that it is indeed a subgroup.

(Nonempty, because it contains at least the identity permutation, and finite because it can only have up to $n!$ elements which is a finite number.)

Now, to show that $H$ is isomorphic to $S_k × S_{n−k}$, I have to define a bijective homomorphism. I don't quite understand what this means. I know that permutations are bijective, but how would this work for $S_k × S_{n−k}$? What does $S_k × S_{n−k}$ actually mean?

Answer 1

In Gallian's "Contemporary Abstract Algebra (Eighth Edition)", there is the following theorem.

Theorem [. . .] (Finite Subgroup Test) Let $H$ be a nonempty finite subset of a group $G$. If $H$ is closed under the operation of $G$, then $H$ is a subgroup of $G$.

The proof is fairly straightforward.

So you are done.

Answer 2

Recall that a permutation of $\{1 \dots n\}$is a bijective map, then for a fixed $\sigma\in H$, the elements of $M$ are all of the form $\sigma(j)$ for $j \in M$. Moreover for the inverse of $\sigma$ in $S_n$ the property $\sigma(j)=j $ holds and this implies $\sigma^{-1} \in H$.

Now, a $\sigma \in H$, restricted to $M$, is a permutation of its $k$ elements. Analogously $\sigma$ acts as a permutation on the complementary of $M$.

Consider the map $H \rightarrow S_k\times S_{n-k}$ that sends $\sigma$ to $(\sigma_{|M},\sigma_{|M^c})$ and try to prove that this is a group isomorphism.