Suppose that $m$ and $n$ are positive integers, and $m<n$.
Define $I:S_m \rightarrow S_n$ as follows:
Given $\alpha \in S_m$, we let
$\hspace{150pt}I(\alpha)(k)=\alpha(k) \hspace{5pt}$ if $k\le m$ and
$\hspace{150pt}I(\alpha)(k)=k \hspace{20pt}$ if $n\ge k >m$.
Show that $I$ is a one-to-one homomorphism, which is not onto. Note: We must first check that $I(\alpha)\in S_n$.
I'm not really sure how to get started on this problem. Can someone point me in the right direction?
On
Hint. $S_m$ is the set of permutations of $\{1,2,\ldots, m\}$ and $S_n$ is the set of permutations of $\{1,2,\ldots,m,m+1, \ldots, n\}$. The question's asking to show that $S_m$ is isomorphic to the subgroup of $S_n$ permuting the first $m$ elements and leaving the rest fixed. Can you show that this subgroup exists? Is the definition of this monomorphism natural?
Outline:
1) Show $I(\alpha)$ is a bijection from $\{1,..,n\}$ to itself. This is simple enough, as $\alpha$ is a bijection of the first $m$ elements to itself, and the rest remain fixed.
2) a) Show $I(\alpha\beta)(k)=I(\alpha)(I(\beta(k))$. Break it into cases as suggested by the definition. This proves $I$ is a homomorphism.
b)Then show $I(\alpha)=I(\beta) \implies \alpha=\beta$. You assume the images are equal at all values of $k\in\{1,...,n\}$, so you need only show $\alpha(k)=\beta(k)$ for $k\le m$.
c) The homomorphism is not onto, because the permutation $(n-1,n)$ has no preimage, for example.