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  1. #1
    Senior Member slevvio's Avatar
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    Hello, I was wondering if I could get some help with this question.

    Let G be the subgroup of  S_4 generated by (1 2),(3 4) and (1 3)(2 4). Let H be the subgroup of S_4 generated by (1 2) and (3 4). Show that H is a normal subgroup of G. Show that H has order 4 and G/H has order 2. Deduce that G has order 8.

    I was wondering if there is a quick way to show that G has order 4 without doing it explicitly? I was also wondering how to show that G/H has order 4. I guess that Lagrange's Theorem proves the last part. Any help with this would be appreciated.

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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by slevvio View Post
    Hello, I was wondering if I could get some help with this question.

    Let G be the subgroup of  S_4 generated by (1 2),(3 4) and (1 3)(2 4). Let H be the subgroup of S_4 generated by (1 2) and (3 4). Show that H is a normal subgroup of G. Show that H has order 4 and G/H has order 2. Deduce that G has order 8.

    I was wondering if there is a quick way to show that G has order 4 without doing it explicitly? I was also wondering how to show that G/H has order 4. I guess that Lagrange's Theorem proves the last part. Any help with this would be appreciated.

    Clearly we have that G\leqslant S_4 and H\leqslant G.

    To see that \left|H\right|=4 we must merely note that \left|(1,2)\right|=\left|(3,4)\right|=\left|(1,2)(  3,4)\right|=2. I leave it to you to prove that \left|G\right|=8 (shouldn't be too hard). It is clear that H\lhd G since \left[G:H\right]=2. So then clearly \left|G/H\right|=\left[G:H\right]=2
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  3. #3
    Senior Member slevvio's Avatar
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    I dont understand how you've shown that H = 4 ?
    But thanks for your reply
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by slevvio View Post
    I dont understand how you've shown that H = 4 ?
    But thanks for your reply
    Clearly we have that H must contain at least \left\{e,(1,2),(3,4),(1,2)(3,4)\right\}. But, it is clear to see that \left((1,2)(3,4)\right)^n=\begin{cases} e & \mbox{if} \quad n\text{ is even} \\ (3,4) & \mbox{if} \quad n\text{ is odd}\end{cases} and similarly for the other two elements and so it remains to check that the multiplication of any combination of them stays within the group. To do this note that \left(1,2\right)^m\left(3,4\right)^n\left((1,2)(3,  4)\right)^\ell=\left(1,2\right)^m(3,4)^n (1,2)^\ell (3,4)^\ell=(1,2)^{m+\ell}(3,4)^{n+\ell}...so
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  5. #5
    MHF Contributor Drexel28's Avatar
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    P.S. H\cong \mathbb{Z}_2\times\mathbb{Z}_2
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  6. #6
    Senior Member slevvio's Avatar
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    Ah I understand now. Obviously those 4 elements are in H, but since they all have order 2 you cannot get any more members when you combine them.

    Thanks for the help
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by slevvio View Post
    Ah I understand now. Obviously those 4 elements are in H, but since they all have order 2 you cannot get any more members when you combine them.

    Thanks for the help
    Mainly because this subgroup happened to be abelian since disjoint cycles commute.
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  8. #8
    Senior Member slevvio's Avatar
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    Is there a way to show that H is normal to G without having to work out the elements and conjugate them one by one?
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  9. #9
    Senior Member slevvio's Avatar
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    Actually what I did was let X = {(1 2) , (3 4)} and let Y = { (1 2), (3 4) , (1 3)(2 4) } and showed  yXy^{-1} = X .
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  10. #10
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by slevvio View Post
    Is there a way to show that H is normal to G without having to work out the elements and conjugate them one by one?
    Yes. I told you that since it has index two it's normal.

    To see this you need to know that a group is normal if every left coset is a right coset. And if a subgroup has index two this is easy to see since we have that the set of left cosets looks like \left\{H,\mathcal{L}\right\} and the set of right cosets looks like \left\{H,\mathcal{R}\right\} and since H\cup\mathcal{L}=G=H\cup\mathcal{R} and H\cap \mathcal{L}=\varnothing=H\cap\mathcal{R} it followst hat \mathcal{L}=\mathcal{R}.
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  11. #11
    Senior Member slevvio's Avatar
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    OK I will think on what you have said

    However the question is asking me to show that | G / H | = 2 and use that to deduce |G| = 8. It's easy to show this the other way round but I can't quite work this out
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  12. #12
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by slevvio View Post
    OK I will think on what you have said

    However the question is asking me to show that | G / H | = 2 and use that to deduce |G| = 8. It's easy to show this the other way round but I can't quite work this out
    If H\lhd G then \left|G/H\right|=\left[G:H\right]
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  13. #13
    Senior Member slevvio's Avatar
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    ah of course. It's very late here in Scotland
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  14. #14
    Senior Member slevvio's Avatar
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    I understand that [G:H] = | G / H | if H is normal to G and I know that |G| = 8.

    But how do I show that [G:H] = 2 only knowing that H has order 4. The question does not ask me to work out |G| but to deduce it from the fact that [G:H] = 2, but how do I show this without showing |G| = 8 first.

    I am trying to show that there must only be 2 distinct cosets of H in G but Im not getting very far
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  15. #15
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by slevvio View Post
    I understand that [G:H] = | G / H | if H is normal to G and I know that |G| = 8.

    But how do I show that [G:H] = 2 only knowing that H has order 4. The question does not ask me to work out |G| but to deduce it from the fact that [G:H] = 2, but how do I show this without showing |G| = 8 first.

    I am trying to show that there must only be 2 distinct cosets of H in G but Im not getting very far
    You have that H\cup H(1,3)(3,4)=G and they are disjoint...so they are the only distinct cosets.
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