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  1. MHB Master
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    #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?

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  3. MHB Master
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    #2
    Quote Originally Posted by mathmari View Post
    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).

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    #3 Thread Author
    Quote Originally Posted by Deveno View Post
    8 is a power of two, that should be a BIG clue, right there.
    What information do we get from that?


    Quote Originally Posted by Deveno View Post
    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?

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    #4
    Quote Originally Posted by mathmari View Post
    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 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"?

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    #5 Thread Author
    Quote Originally Posted by Deveno View Post
    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?


    Quote Originally Posted by Deveno View Post
    do you think you can take them "all the way"?
    What do you mean?

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    #6
    Quote Originally Posted by mathmari View Post
    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 Quote:
    What do you mean?
    I mean keep going.

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    #7 Thread Author
    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?
    Last edited by mathmari; March 30th, 2016 at 09:04.

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    #8
    Quote Originally Posted by mathmari View Post
    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.

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    #9 Thread Author
    Quote Originally Posted by Deveno View Post
    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!!

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