# Thread: Normal series and composition series

1. 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? But how can we find all the composition series?   Reply With Quote

2.

3. Originally Posted by 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? But how can we find all the composition series? 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).  Reply With Quote Originally Posted by Deveno 8 is a power of two, that should be a BIG clue, right there.
What information do we get from that?  Originally Posted by Deveno 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?   Reply With Quote

5. Originally Posted by mathmari What information do we get from that? The only ways we can go down is by "halfsies". (All the factor groups in a composition series will have order 2). Quote:
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? Mhm...so each of these looks promising as the start of a composition series...do you think you can take them "all the way"?  Reply With Quote Originally Posted by Deveno 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? 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? Why isn't this a composition series?  Originally Posted by Deveno do you think you can take them "all the way"?
What do you mean?   Reply With Quote

7. Originally Posted by mathmari When we had a group of order a power of $3$, all the factor groups in a composition series would have order $3$, right? 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? Why isn't this a composition series? Who said it wasn't? Quote:
What do you mean? I mean keep going.  Reply With Quote

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? How do we know if we have found all composition series?   Reply With Quote

9. Originally Posted by 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? How do we know if we have found all composition series? 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.  Reply With Quote Originally Posted by Deveno 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!!   Reply With Quote

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