(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

For n>1, show that the subgroup H of [itex] S_n [/itex] (the symmetric group on n-letters) consisting of permutations that fix 1 is isomorphic to [itex] S_{n-1} [/itex]. Prove that there are no proper subgroups of [itex] S_n [/itex] that properly contain H.

3. The attempt at a solution

The first part is fairly straightforward. Using cycle notation, we ignore the trivial cycle on 1, then just reduce all other components by 1. It can be shown to be bijective hence giving the desired isomorphism.

Now for the second part, I figure this simply amounts to showing that there is no group N such that [itex] S_{n-1} \subsetneq N \subsetneq S_n [/itex]. My first thought is to attack the problem via contradiction. If such a subgroup existed, it would be necessary that [itex] (n-1)! \big| |N| \big| n! [/itex]. Unfortunately, I cannot find a reason this can't be true, except when n is prime.

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# Homework Help: Intermediate subgroups between symmetric groups

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