For each $n$ determine whether $S_n$ (1) and $D_n$ (2) are simple or not.

Question

For each n determine whether $S_n$ (1) and $D_n$ (2) are simple or not.

So I know that a group is called a simple group if that group has no proper nontrivial normal subgroups.

(1) First of all I need to find if $S_n$ has normal subgroups. Am I right?

In one example I found that $A_n$ is normal subgroup of $S_n$. How should I check it? $|S_n|=n!$ and $|A_n|=n!/2$ so $[S_n:A_n]=2$ I think $A_n$ is a subgroup if index $2$ in $S_n$. So $S_n$ is not simple?

(2) I am not sure if I was right in (1), but with this I have no idea what to do. I only know that $D_n=\{1,\alpha,\alpha^2,...,\alpha^{n-1},\beta,\alpha\beta,...,\alpha^{n-1}\beta |\alpha^n=\beta^2=1, \beta\alpha=\alpha^{-1}\beta \}$.

Answer 1

You have the correct idea.

For $n>2$, $S_n$ has a proper normal subgroup $A_n$. So $S_n$ is not simple. However, $S_2=Z_2$ is simple.

For dihedral group, consider the subgroup $\langle \alpha \rangle$. Since $|\alpha|=n$, $|D_n:\langle \alpha \rangle|=2$, hence $\langle \alpha \rangle$ is a proper normal subgroup of $D_n$ when $n>2$. Similarly, $D_2=Z_2$ is simple.