# Derived subgroup of $S_n$ and $D_n$

I know that Derived/Commutator subgroup of $S_3$ is $A_3$ and commutator subgroup of $D_4$ is cyclic of order $2$.

But What about derived groups of $S_n$ and $D_n$?

How can I calculate them?

• (This would probably be better as two separate questions.) The commutator subgroup of $S_n$ is always $A_n$. The commutator subgroup of $D_{2n}$ is $\langle r^2 \rangle$ where $r$ is the generator of order $n$. – George V. Williams Jun 28 '15 at 20:14

1. Symmetric Groups

Let $G=S_n$. Then as $S_n/A_n \cong \Bbb{Z_2} \implies G' \subseteq A_n$.

Claim- $G'=A_n$.

Proof- Let $\sigma \in A_n$. Then by a standard resule, we know $A_n$ is generated by $3$-cycles, hence cycle $\sigma= t_1t_2\dots t_n$ where $t_i's$ are $3$- cycles. Now let any arbitrary three cycle $(abc)$. It can be rewritten as $(abc)=(ab)(ac)(ba)(ca)=(ab)(ac)(ab)^{-1}(ac)^{-1} \in G'$.
So we have proved $\sigma \in G'$. Thus, $S_n'=A_n \hspace{9cm}\blacksquare$

Now I encourage you to find $A_n'$ for all $n> 1$.

HINT- $$A_2'=e \\ A_3'=e\\ A_4'=V_4\\$$

But $\hspace{7.3cm}A_5'=G(n)$ $\hspace{2cm} \forall\ n\ge 5$.

Find $G(n)$ for all $n\ge 5$

NOTE- $A_n$ is simple for $n\ge 5$.

1. Dihedral Groups-

$G=D_n=\{1,r,r^2,\dots r^{n-1}, s, sr, \dots sr^{n-1}\}$, where $s^2=1$ and $srs^{-1}=r^{-1}$. Here $r^i$ are rotations and $r^is$ are reflections.

Now It is easy to see that $r^2=rsr^{-1}s^{-1}$. Now check that $r^{2k}=r^ksr^{-k}s^{-1}$, which implies $(r^2) \le G'$.

Claim- $D_n'=(r^2)$, i.e. $[a,b]\in (r^2)\ \forall\ a,b\in D_n$.

The only interesting case is when one of $a$ and $b$ is a reflection and other is a rotation.

So let $[a,b]=[r^i,r^ks]=aba^{-1}b=r^ir^k(s(r^{-i}r^k)s^{-1})=r^{i+k}r^{-(k-i)}=r^{2i}\in (r^2)$.