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Math Help - Normal Subgroup

  1. #1
    ux0
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    Normal Subgroup

    Define W= <(1 2)(3 4)>, the cyclic subgroup of S_4 generated by (1 2)(3 4). Show that W is a normal subgroup of V, where V ={(1), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}, but that W is not a normal subgroup of  S_4 . Conclude that normality is not transitive: K is normal to H, and H is normal to G does not imply K is normal to G.
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    Quote Originally Posted by ux0 View Post
    Define W= <(1 2)(3 4)>, the cyclic subgroup of S_4 generated by (1 2)(3 4). Show that W is a normal subgroup of V, where V ={(1), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}, but that W is not a normal subgroup of  S_4 . Conclude that normality is not transitive: K is normal to H, and H is normal to G does not imply K is normal to G.
    Hi!

    V is also otherwise known as the Klein-four-group(!)

    We know that W is normal if \forall g \in G, w \in W , gwg^{-1} \in W

    However since the elements of W are only (2)(2) circles and so are the elements of V, and by the fact that \forall \sigma,\rho \in S_n such that  \rho = (\rho(1), \rho(2),...\rho(n)), we know that: \sigma \rho \sigma^{-1} = (\sigma(\rho(1)), \sigma(\rho(2)),...,\sigma(\rho(n)))

    Specifically, this gives us that the circle-form of \sigma \rho \sigma^{-1} is determined solely by that of \rho. This gives us that W is normal since the result is always a 2-2 circle and all of them are contained in the klein group.

    To do the second part simply choose some other element \xi \in S_n and show that \xi \rho \xi^{-1} \notin W
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    Quote Originally Posted by Defunkt View Post
    Hi!

    However since the elements of W are only (2)(2) circles and so are the elements of V,
    Just to clarify some things, this means that the elements of W are of order 2 and cyclic?.. I'm not really sure what you mean by circles we haven't discussed this in these terms in class.
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    Quote Originally Posted by ux0 View Post
    Define W= <(1 2)(3 4)>, the cyclic subgroup of S_4 generated by (1 2)(3 4). Show that W is a normal subgroup of V, where V ={(1), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}, but that W is not a normal subgroup of  S_4 . Conclude that normality is not transitive: K is normal to H, and H is normal to G does not imply K is normal to G.

    What've you done so far? Where're you stuck? V is what kind of group? Thus any of its subgroup is...
    What about conjugating (12)(34)\,\,by\,\,(13)?

    Tonio
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    Quote Originally Posted by tonio View Post
    What've you done so far?
    Not much we are using "First course in Abstract Algebra, by Rotman, we are are Homomoprhism, all we have covered is some set theory, number theory, and a few groups.

    Quote Originally Posted by tonio View Post
    Where're you stuck?
    Im stuck on this proof because i don't understand what he means by (2)(2) circles, or later in the proof when he says 2-2 circles.... Is he saying that V has index 2 in a group G? where S_4 = G.



    Quote Originally Posted by tonio View Post
    V is what kind of group? Thus any of its subgroup is...
    I know V is a normal subgroup of S_4 ... DOes that imply that if H is a subgroup of index 2 in a group G, then H is a normal subgroup of G?


    Quote Originally Posted by tonio View Post
    What about conjugating (12)(34)\,\,by\,\,(13)?
    (13) (12)(34) (31) = (3 2) (3 4) ??

    i'm not really sure on how to do that?
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  6. #6
    MHF Contributor Swlabr's Avatar
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    I'm going to start from the beginning, because I can't understand the above post...this may not actually be helpful, but I can but try...

    Quote Originally Posted by ux0 View Post
    Define W= <(1 2)(3 4)>, the cyclic subgroup of S_4 generated by (1 2)(3 4). Show that W is a normal subgroup of V, where V ={(1), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}, but that W is not a normal subgroup of  S_4 . Conclude that normality is not transitive: K is normal to H, and H is normal to G does not imply K is normal to G.
    Firstly, note that V has order 4 and so is abelian (all groups of order 4 are abelian - there are only two of them). Thus, W \lhd V, because subgroups of abelian groups are always normal.

    However, W \not\lhd S_4 because you can find an element \sigma \in S_4 such that \sigma^{-1}(12)(34)\sigma \notin W and so W cannot be normal in S_4.

    Such a \sigma is \sigma=(13). (13)(12)(34)(13)=(14)(23) \notin W. Do you know how to perform such calculations? You go from left to right, following where each number sends you. For instance, start with 1. 1 \rightarrow 3 \rightarrow 4. Then plugging in 4 we get 4 \rightarrow 3 \rightarrow 1, and so we have found the first part of the cycle, (14). Then do the same with 2, and you find it goes to 3, and then we know 3 must go to 2.
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    Just to clarify, a 2-2 circle is a permutation of the form (a b)(c d). What I said was that the result of \sigma \rho \sigma^{-1} where \rho is a 2-2 circle will always be a 2-2 circle. However, all 2-2 circles are contained in V, so this gives that W is normal. Swlabr's way is alot easier to understand, though.
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  8. #8
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    Quote Originally Posted by Swlabr View Post
    .
    \sigma=(13). (13)(12)(34)(13)=(14)(23) \notin W.
    Shouldn't \sigma^{-1}=(31) therefore making the (31)(12)(34)(13)=(12)(14)=(2 4) \notin W???
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  9. #9
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by ux0 View Post
    Shouldn't \sigma^{-1}=(31) therefore making the (31)(12)(34)(13)=(12)(14)=(2 4) \notin W???
    Yes and no to your first question: (13)=(31) (put them both into a proper permutation to see why - they both move 1 to 3 and 3 to 1 and leave everything else unchanged).

    No to your second part - your calculation is incorrect. For starters, you should always end up with disjoint cycles after performing the multiplication the first time. You round off the brackets if and only if you get back to the beginning (as in, you go a_1 \rightarrow a_2 \rightarrow \ldots \rightarrow a_i \rightarrow a_1 and so your cycle will be (a_1 a_2 \ldots a_i)).
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