Proving that two permutation groups are isomorphic Here's the statement to prove: Let $n,m$ be two positive integers with $m≤n$. 

Prove that $S_m$ is isomorphic to a subgroup of $S_n$, where $S_n$ is the collection of all permutations of the set $n$, nonempty, and $S_m$ is the collection of all permutations of the set $m$, nonempty. 

I am quite lost here. Is it sufficient to pick a subset of $S_n$, with $n=m$, which has order $m!$? Which would mean that $S_n$, with $n=m$, and $S_m$ would be isomorphic to $Z6.$ 
 A: For each $\sigma \in S_m$, define a permuation $\hat{\sigma}$ on $\{1,2,\ldots, n\}$ by
$$
\hat{\sigma}(j) = \begin{cases}
\sigma(j) &: j\leq m \\
j &: j > m
\end{cases}
$$
Check that $\hat{\sigma} \in S_n$ and that
$$
\sigma \mapsto \hat{\sigma}
$$
is an isomorphism.
A: Remember what it means to be isomorphic: a bijective homomorphism. You're on the right track: if $S_m$ is bijective to a subgroup of $S_n$, that subgroup must have order $|S_m|=m!$. 
However, you need a little bit more: that subgroup must also have the same group structure (it is a bijective homomorphism, not just a bijection!): loosely speaking, you "pair up" each element of $S_m$ with an element of the subgroup of $S_n$ (bijection), in a way that multiplication looks the same in both groups (homomorphism). For a more rigorous definition, check your textbook.
As you said, $S_m$ is the permutations of the set $\{1, \ldots, m\}$, and $S_n$ is the permutations of the set $\{1, \ldots, m, \ldots, n\}$. Do you see a natural way to associate each permutation of $S_m$ with one in $S_n$?
A: If you pick a subset of m symbols what would be the order of the permutation group? Consider a permuation on those m symbols. What are the possible ways a permutation can map those m symbols to itself? So, what is the number of elements of the set of permutations?  Then use the properties of permutations to show that the set of permutations is a group?
