Suppose that $X$ is a nonempty subset of a set $Y$. Show that $S_X$ is isomorphic to a subgroup of $S_Y$

Question

Suppose that $X$ is a nonempty subset of a set $Y$.
Show that $S_X$ is isomorphic to a subgroup of $S_Y$.

Updated idea:

Define $f:S_X\rightarrow f(S_X)$ by $$f(\sigma x)=\sigma x,\forall x\in X$$ $$f(\sigma y)=y,\forall y\in Y-X$$ where $\sigma \in S_X$

Let $\sigma,\tau \in S_X$
Suppose $f(\sigma)=f(\tau)$
Then $\sigma x=\tau x,\forall x\in X$
and $\sigma y=\tau y=y,\forall y\in Y-X$
Hence $\sigma=\tau$ and $f$ is injective.

Next it is clearly that $f$ is surjective

Since $f(\sigma \circ \tau)=f(\sigma)\circ f(\tau)$
$f$ is homomorphism.

Note that $f(S_X)=\{\alpha\in S_X |\alpha(y)=y,\forall y\in Y-X\}$
(i)$1\in f(S_X)$
(ii)$\alpha,\beta \in f(S_X)\implies \alpha\circ\beta^{-1}y=y,\forall y\in Y-X$
Hence,$f(S_X)\leq S_Y$

Answer 1

Take a permutation $\sigma$ on $X$. Now extend $\sigma$ to a permutation $f(\sigma)$ on $Y$ by setting $f(\sigma)(y) = y$ for all $y \in Y - X$. Then $f$ is an injective morphism and $f(S_X)$ is a subgroup of $S_Y$.