I was working on an algebra exercise which states the following:
Show that the symmetric group $S_n$ is a maximal subgroup of $S_{n+1}$. (Hint: Show that if $g \in S_{n+1} \setminus S_n$, then $S_{n+1} = S_n \cup S_n g S_n$.)
I'm not sure how to use this hint to solve this exercise; I'm not really sure how to being approaching this problem if I chose not to use the hint. Does anyone have any suggestions?
On
I suppose you take $\;S_n:=\left\{\,\sigma\in S_{n+1}\;/\;\sigma(n+1)=n+1\,\right\}\le S_{n+1}\;$ ...though this is completely unimportant.
Suppose then that there exists $\;H\lneqq S_{n+1}\;$ s.t. $\;S_n\lneqq H\implies \exists\,g\in H\setminus S_n\;$ , and thus using the hint we get that
$$S_{n+1}=S_n\cup S_ngS_n\le S_n\cup S_n HS_n\le S_n\cup H=H\;,\;\;\text{ contradiction.}$$
If you prove the hint, then any group containing $S_n$ and some $g\in S_{n+1}/S_n$ then it must contain $S_ngS_n$ and therefore $S_n\cup S_ngS_n=S_{n+1}$.
The obvious approach to proving this, I think, is to assume $S_n< H\le S_{n+1}$ and show that any $h\in S_{n+1}$ is in $H$. To do this you would take $g\in H\setminus S_n$ and write $h$ in terms of $g$ and elements in $S_n$, so you would almost prove the hint anyway.