Isomorphism between $S_n$ and a subgroup of $S_{n+1}$ Let $S_n$ be the symmetric group on $n$ letters. Now, I wish to show that $S_n$ is isomorphic to the subgroup of all the elements of $S_{n+1}$  that leaves $n+1$ fixed. 
Let $\sigma \in S_n$. Define the function $\phi:S_n \rightarrow S_{n+1}$ as follows:
$$\phi(\sigma) (m)=\begin{cases} \sigma(m)  \ & m \le n \\ n+1 &m=n+1 \end{cases}$$
Clearly, $\phi$ is both one-to-one and onto. Let $\tau \in S_n$, too. It remains to prove that $\phi(\sigma  \circ \tau)=\phi(\sigma) \ \circ \phi(\tau)$. But this is obvious, since:
$$\phi(\sigma \circ \tau)(m)= \begin{cases} (\sigma \circ \tau)(m) & m \le n \\ n+1  & m=n+1\end{cases}=\phi(\sigma) \ \circ \phi(\tau)$$
Hence, we are done. 
Does the presentation make sense? In particular, I want to check if my choice of notations are appropriate.
 A: Your $\phi$ is not onto: not every permutation on $n+1$ letters fixes $n+1$. But you don't need that for your aim: injectivity and composition preservation are enough to embed $S_n$ into $S_{n+1}$, meaning that the former is isomorphic to a subgroup of the latter.
Anyway, your strategy is correct. More generally, let $X$ be a set and $Y \subseteq X$. For a given bijection $u$ on $X \setminus Y$, any bijection $\alpha$ on $Y$ can be extended to a bijection $f_u(\alpha)$ on $X$ by:
\begin{alignat}{1}
&f_u(\alpha)_{|Y}:=\alpha \\
&f_u(\alpha)_{|X \setminus Y}:=u \\
\tag 1
\end{alignat}
Note that $f_u(\alpha)=f_u(\beta) \Rightarrow f_u(\alpha)_{|Y}=f_u(\beta)_{|Y} \Rightarrow \alpha=\beta$, so that $f_u$ is injective for all $u \in \operatorname{Sym}(X \setminus Y)$. Moreover,
$$f_u(\alpha\beta)=f_u(\alpha)f_u(\beta) \iff u=\iota_{X \setminus Y} \tag 2$$
($\iota_B$ is the identical map on the set $B$: $\iota_B(b)=b$ for every $b \in B$.) Therefore, if we take $u=\iota_{X \setminus Y}$, we have $f_u : {\rm{Sym}}(Y) \hookrightarrow {\rm{Sym}}(X)$. Your case is the particular one with $X=\{1,\dots,n+1\}$ and $Y=\{1,\dots,n\}$.
