MHB Normal series and composition series

[mathmari]

Hey!

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)

 

[Deveno]

Hey!

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)

8 is a power of two, that should be a BIG clue, right there.

So find the "other" subgroups of order 4 (there are two more).

 

[mathmari]

8 is a power of two, that should be a BIG clue, right there.

What information do we get from that? (Wondering)

So find the "other" subgroups of order 4 (there are two more).

The other two subgroups of order $4$ are $\langle a, s^2\rangle=\{a, s^2, as^2, 1\}$ and $\langle s^2, as\rangle=\{s^2, as, as^3, 1\}$, right? (Wondering)

 

[Deveno]

What information do we get from that? (Wondering)

The only ways we can go down is by "halfsies". (All the factor groups in a composition series will have order 2).

The other two subgroups of order $4$ are $\langle a, s^2\rangle=\{a, s^2, as^2, 1\}$ and $\langle s^2, as\rangle=\{s^2, as, as^3, 1\}$, right? (Wondering)

Mhm...so each of these looks promising as the start of a composition series...do you think you can take them "all the way"?

 

[mathmari]

The only ways we can go down is by "halfsies". (All the factor groups in a composition series will have order 2).

When we had a group of order a power of $3$, all the factor groups in a composition series would have order $3$, right? (Wondering)

In post #1 all the factor groups in $1\leq \langle s^2\rangle \leq \langle s\rangle \leq D_4$ have order $2$, right? (Wondering)

Why isn't this a composition series? (Wondering)

do you think you can take them "all the way"?

What do you mean? (Wondering)

 

[Deveno]

When we had a group of order a power of $3$, all the factor groups in a composition series would have order $3$, right? (Wondering)

In post #1 all the factor groups in $1\leq \langle s^2\rangle \leq \langle s\rangle \leq D_4$ have order $2$, right? (Wondering)

Why isn't this a composition series? (Wondering)

Who said it wasn't?

What do you mean? (Wondering)

I mean keep going.

 

[mathmari]

So are the following some the composition series?

$$1\leq \langle s^2\rangle \leq \langle s\rangle \leq D_4 \\ 1\leq \langle s^2\rangle \leq \langle s^2, a\rangle \leq D_4 \\ 1\leq \langle s^2\rangle \leq \langle s^2, as\rangle \leq D_4 $$

Other subgroups of order $2$ are $\langle a\rangle=\{a, a^2=1\}$, $\langle as^2\rangle=\{as^2, 1\}$, $\langle as^3\rangle=\{as^3, 1\}$, $\langle as\rangle=\{as, 1\}$.

Therefore, other composition series are the following:

$$1\leq \langle a\rangle \leq \langle s^2, a\rangle \leq D_4 \\ 1\leq \langle as^2\rangle \leq \langle s^2, a\rangle \leq D_4 \\ 1\leq \langle as^3\rangle \leq \langle s^2, as\rangle \leq D_4 \\ 1\leq \langle as\rangle \leq \langle s^2, as\rangle \leq D_4$$
right? (Wondering)

How do we know if we have found all composition series? (Wondering)

 

[Deveno]

So are the following some the composition series?

$$1\leq \langle s^2\rangle \leq \langle s\rangle \leq D_4 \\ 1\leq \langle s^2\rangle \leq \langle s^2, a\rangle \leq D_4 \\ 1\leq \langle s^2\rangle \leq \langle s^2, as\rangle \leq D_4 $$

Other subgroups of order $2$ are $\langle a\rangle=\{a, a^2=1\}$, $\langle as^2\rangle=\{as^2, 1\}$, $\langle as^3\rangle=\{as^3, 1\}$, $\langle as\rangle=\{as, 1\}$.

Therefore, other composition series are the following:

$$1\leq \langle a\rangle \leq \langle s^2, a\rangle \leq D_4 \\ 1\leq \langle as^2\rangle \leq \langle s^2, a\rangle \leq D_4 \\ 1\leq \langle as^3\rangle \leq \langle s^2, as\rangle \leq D_4 \\ 1\leq \langle as\rangle \leq \langle s^2, as\rangle \leq D_4$$
right? (Wondering)

How do we know if we have found all composition series? (Wondering)

Basically, to find them all, all you need to do is find every subgroup of $D_4$ of index 2, and every subgroup of those subgroups of index 2.

There are 3 subgroups of index 2, one is cyclic, and two are isomorphic to $V$, the Klein 4-group.

There is only one composition series for the cyclic subgroup of index 2.

Each of the other subgroups isomorphic to $V$ yield 3 composition series each (one for each element of order 2 they contain).

That gives 7 in all. You've listed 7, so those are them.

 

[mathmari]

Basically, to find them all, all you need to do is find every subgroup of $D_4$ of index 2, and every subgroup of those subgroups of index 2.

There are 3 subgroups of index 2, one is cyclic, and two are isomorphic to $V$, the Klein 4-group.

There is only one composition series for the cyclic subgroup of index 2.

Each of the other subgroups isomorphic to $V$ yield 3 composition series each (one for each element of order 2 they contain).

That gives 7 in all. You've listed 7, so those are them.

Great! Thank you very much! (Happy)