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First of all I don't know how to write with correct notation in this site, sorry about that.

Second, I am a beginner with group theory so I am trying to solve this exercise (Fraleigh).

Until now I have a function that maps a permutation in $S_7$ to $S_8$ like this:

Suppose $\sigma$ , any permutation in $S_7$.

$5$ if $x=5$

$\sigma(x)$ if $\sigma(x)\neq 5$

Here I have the first problem, because I have two cells free: $8$ and the element that is sent to $5$ in $S_7$, so I maps like this:

$\sigma(5)$ if $\sigma(x)=5$

$8$ if $x=8$

So I would appreciate your help in this exercise, I don't know if this function is correct and how to prove if in fact is an isomorphism.

Kan't
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Jhon C
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2 Answers2

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I'm sure this has been answered here before but, since I can't find a duplicate, I'll give a CW answer.

There's nothing special about $7$, $8$, and $5$ here.

Theorem: Let $n\in \Bbb N$ and $m\in\{1,\dots, n+1\}$. Then $S_n$ is isomorphic to the subgroup $H$ of $S_{n+1}$ whose elements fix $m$.

Proof: We may assume, by reindexing if necessary, that $m=n+1$.

We have that

$$\begin{align} \varphi: H&\to S_n,\\ \sigma &\mapsto \sigma_H, \end{align}$$

where

$$\sigma_H(x)=\sigma(x)$$

for $x\in \{1,\dots, n\}$, is an isomorphism, since it is clearly a bijection and

$$\begin{align} (\varphi(\tau\rho))(x)&=(\tau\rho)_H(x)\\ &= (\tau\rho)(x)\\ &= \tau(\rho(x))\\ &=\tau(\rho_H(x))\\ &=\tau_H(\rho_H(x))\\ &=(\varphi(\tau))((\varphi(\rho))(x))\\ &=(\varphi(\tau)\varphi(\rho))(x) \end{align}$$

for all $x\in \{1,\dots, n\}$ and all $\tau,\rho\in H$, so that

$$\varphi(\tau\rho)=\varphi(\tau)\varphi(\rho).\,\square$$

Shaun
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  • I'm trying to understand your answer but I'm really confuse. The permutation (1, 5)(2, 3, 4)(6, 7) in $S_7$, maps 1 to 5. When I apply your function, 5 maps to 5, 2 maps to 3, 3 to 4, 4 to 2, 6 to 7 and 7 to 6. But now, 1 is empty, because in $S_7$ the image of 1 was 5, and same for 8. If I map 8 to itself for every permutation, the subgroup of $S_8$ has 6! elements and not 7! So, those are the real doubts: 1 (in $S_8$) should be go to the image of 5 (in $S_7$)? or should be go to 8? What if my permutation in $S_7$, 5 maps to 5? I would really appreciate your help. – Jhon C Jul 13 '22 at 02:08
  • Well spotted, @JhonC! I have repaired the damage. – Shaun Jul 13 '22 at 12:07
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For every $a\in A:=\{1,\dots,n\}$, set $B:=A\setminus\{a\}$. Then, $\operatorname{Stab}(a)\stackrel{\varphi}{\cong} S_B$ via $\sigma\mapsto\varphi(\sigma):=\sigma_{|B}$. In fact, firstly, $\sigma_{|B}\in S_B$ because $\sigma \in\operatorname {Stab}(a)$ (good definition). Moreover:

  • injectivity: $\varphi(\sigma)=\varphi(\tau)\Longrightarrow \sigma_{|B}=\tau_{|B}\stackrel{\sigma(a)=a=\tau(a)}{\Longrightarrow}\sigma=\tau$;
  • surjectivity: for any given $f\in S_B$, define $\sigma\in S_n$ by $(\sigma(a):=a) \wedge (\sigma_{|B}:=f)$; therefore, $\sigma\in\operatorname{Stab}(a)$ and $\varphi(\sigma)=f$;
  • operation-preserving: for $b\in B$: \begin{alignat}{1} (\sigma\tau)(b) &= \sigma(\tau(b)) \\ &\stackrel{b\in B}{=}\sigma(\tau_{|B}(b)) \\ &\stackrel{\tau_{|B}(b)\in B}{=}\sigma_{|B}(\tau_{|B}(b)) \\ &=(\sigma_{|B}\tau_{|B})(b) \\ \end{alignat} On the other hand, $b\in B\Longrightarrow (\sigma\tau)(b)=(\sigma\tau)_{|B}(b)$. Therefore: $$\varphi(\sigma\tau)=(\sigma\tau)_{|B}=\sigma_{|B}\tau_{|B}=\varphi(\sigma)\varphi(\tau)$$ Now, note that $|B|=n-1$, and hence $S_B\stackrel{\phi}{\cong} S_{n-1}$ for $\phi(f):=\mathscr gf\mathscr g^{-1}$, where $\mathscr g\colon B\longrightarrow \{1,\dots,n-1\}$ is any bijection. So, finally, for every $a\in A$: $$\operatorname{Stab}(a)\stackrel{\phi\varphi}{\cong} S_{n-1}$$ Yours is just the particular case $n=8$ and $a=5$.