I am stucky in a problem about symmetric groups.
If $k+l=n$, $k\neq l$ and we consider the natural embedding in $S_n$, how can I show that this subgroup $S_k \times S_l$ is a maximal subgroup of $S_n$?
First doubt: can someone explain me that natural embedding?
I think I need to use group action to prove that but I really have no idea how to start it.
Can someone help me with some hints?
Thank you so much, everyone! :)
Hint: A permutation that is not in the subgroup isomorphic to $S_k\times S_l$ will send at least one element among the first $k$ to one among the last $l$, and also vice versa. Try to prove that by combining this one permutation and arbitrary elements of $S_k\times S_l$ you can make any permutation whatsoever.
Thinking of it, the result does not seem true in general. Add to $S_2\times S_2$ inside $S_4$ the order-reversing permutation, and they only generate a dihedral group of order $8$, a proper subgroup of $S_4$. Oops, I missed the condition $k\neq l$ in the question. So now you need to think of exploiting that condition.