$S_7$ is isomorphic to subgroup of all those elements of of $S_8$ that sends $8$ to $8$

Question

I have to show that $S_7$ is isomorphic to the subgroup of all those elements of $S_8$ that sends $8$ to $8$.

My attempt: let $H$ be subgroup of $S_8$ such that all elements of $H$ sends $8$ to $8$ that is, all elements of $H$ fix the symbol $8$. I had defined map $f$ from $H$ to $S_7$ such that if $\alpha\in H$ then $f(\alpha)= \beta$ where $\beta= \left(\begin{smallmatrix} 1 & 2 & \cdots & 7\\ \alpha(1) & \alpha(2) & \cdots & \alpha(7)\end{smallmatrix}\right)$

I had already proved that $f$ is bijective. But I am not able to show $f$ is homomorphism (operation preserving) map. i.e. how to show $f(\alpha_1\alpha_2)= f(\alpha_1)f(\alpha_2)$

Please help.

Answer 1

$\newcommand{\l}[1]{\{1,2,\ldots,#1\}}$

First, I make the map explicit.

Let $\phi \in S_8$ be a map sending $8$ to $8$. We define $$f(\phi) : \l{7} \to \l{7} \quad ; \quad (f(\phi))(x) = \phi(x)$$

Now we have to prove that $f(\phi)\circ f(\psi) = f (\phi \circ \psi)$ for every $\phi,\psi \in S_8$ which send $8$ to $8$.

To do this, simply check what each side is, evaluated at some $x \in \l{7}$.

By definition, we have \begin{align} f(\phi \circ \psi)(x) = &\phi \circ \psi(x) &\text{definition of $f$} \\ =& \phi(\psi(x)) &\text{definition of composition} \\ = &\phi((f(\psi))(x)) &\text{definition of $f$, inner bracket} \\ = &\big(f(\phi)\big) \big((f(\psi))(x)\big) &\text{see * below} \\ =& [f(\phi) \circ f(\psi)](x) &\text{definition of composition} \end{align}

for all $x$, hence $f(\phi \circ \psi) = f(\phi) \circ f(\psi)$. We get the homomorphism property.

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(*) The way this is is being applied is that $\phi(y) = (f(\phi)) (y)$ for any $ y$ by definition of $f$. Now simply put $y = (f(\psi))(x)$ to get that statement.

Answer 2

In general, 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,

$$f_u(\alpha\beta)=f_u(\alpha)f_u(\beta) \iff u=\iota_{X \setminus Y} \tag 2$$

($\iota_A$ is the identical map on the set $A$: $\iota_A(a)=a$ for every $a \in A$.) Therefore, if we take $u=\iota_{X \setminus Y}$, we have $f_u : {\rm{Sym}}(Y) \hookrightarrow {\rm{Sym}}(X)$. Now, take $Y:=\{1,\dots,7\}\subseteq X:=\{1,\dots,8\}$, and you're done.