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Let $S_n$ denote the symmetric group on $\{1,\ldots,n\}$. Let $M$ be the subgroup $\{\sigma \in S_n \mid \sigma(1) = 1\}$. Show that $M$ is a maximal subgroup of $S_n$.

Here is what I've come up with so far:

Suppose we have a subgroup $H$ of $S_n$ such that $M \subseteq H \subseteq S_n$ . We must show that $H = M$ or $H = S_n$. Thus, it suffices to show that if $H$ is a subgroup of $S_n$ that contains $M$ together with at least one element of $S_n$ that is not in $M$, then $H$ must be all of $S_n$.

To this point, suppose $H$ contains $M$ and a permutation $\beta$ such that $\beta(1) \neq 1$.

Is there a reason why $H$ must then be all of $S_n$ ? I'm hoping there's a clever way to see why that must be true, and I need some help hashing this out.

Thanks!

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1 Answer 1

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It is enough to show that $H$ contains all transpositions of the form $(1,j)$, $j=2,\ldots,n$.

Let $k\neq 1$ be such that $\beta(k)=1$, and let $\beta(1)=r\neq 1$. Multiplying $\beta$ on the left by $(r,k)\in M$ we get a permutation that sends $1$ to $k$ and $k$ to $1$. Thus, $(r,k)\beta = (1,k)\tau$ for some $\tau$ that is disjoint from $\{1,k\}$, so $\tau\in M$. Therefore, $(1,k)\in H$.

Now conjugating with $(k,j)\in M$, $2\leq j\leq n$, we get $(1,j)\in H$, proving that $H$ contains all transpositions of the form $(1,j)$, and hence contains $S_n$.

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  • $\begingroup$ Brilliant. Thank you. $\endgroup$ Commented Dec 11, 2019 at 22:05
  • $\begingroup$ This proof assumes that $r\neq k$. I think that it should be mentioned that if $r=k$, then $\beta$ is already of the form $(1,k)\tau$, with $\tau\in M$. $\endgroup$
    – Adren
    Commented May 11, 2021 at 9:37
  • $\begingroup$ @Adren: There is no need to separate cases. If $r=k$, then $(r,k)$ is the identity. So there is no such assumption. $\endgroup$ Commented May 11, 2021 at 10:22
  • $\begingroup$ I agree. It's only a question of notation : some people consider that the symbol $(i,j)$ denotes a transposition (with the implicit assumption that $i\neq j$). $\endgroup$
    – Adren
    Commented May 11, 2021 at 10:28

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