- #1

- 5,053

- 7

I want to find a normal series of $D_4$ and all the composition series for $D_4$.

I have done the following:

$D_4=\langle a , s\mid s^4=1=a^2, asa=s^{-1}\rangle$

A subgroup of $D_4$ is $\langle s\rangle=\{s, s^2, s^3, s^4=1\}$, that is normal in $D_4$, since $[D_4:\langle s\rangle]=\frac{|D_4|}{|\langle s\rangle|}=2$.

Therefore a normal series of $D_4$ is $$D_4\geq \langle s\rangle \geq 1$$

A composition series for $D_4$ is $$1=S_0\leq S_1\leq S_2\leq \dots \leq S_k=D_4$$ where $S_i\trianglelefteq S_{i+1}$ and $S_{i+1}/S_i$ is a simple group.

A subgroup of $D_4$ is $\langle s\rangle$. We have that $|D_4/\langle s\rangle |=\frac{|D_4|}{\langle s\rangle |}=2$.

Since the order of the quotient is the prime $2$, the group $D_4/\langle s\rangle$ is simple, and $\langle s\rangle \trianglelefteq D_4$.

A subgroup of $\langle s\rangle$ is $\langle s^2\rangle=\{s^2, s^4=1\}$. We have that $|\langle s\rangle/\langle s^2\rangle |=\frac{|\langle s\rangle |}{\langle s^2\rangle |}=2$.

Since the order of the quotient is the prime $2$, the group $\langle s\rangle /\langle s^2\rangle$ is simple, and $\langle s^2\rangle \trianglelefteq \langle s\rangle$.

We have that $|\langle s^2\rangle /1|=|\langle s^2\rangle |=2$, so $\langle s^2\rangle$ is simple and $1 \trianglelefteq \langle s^2\rangle$.

Therefore, a composition series for $D_4$ is $$1\leq \langle s^2\rangle \leq \langle s\rangle \leq D_4$$ right? (Wondering)

But how can we find all the composition series? (Wondering)