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Math Help - Index/Isomorphism/Quotient Group answer check

  1. #1
    Super Member Showcase_22's Avatar
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    Index/Isomorphism/Quotient Group answer check

    Let H be a subgroup of a group G and let X be the set of left cosets xH of H in G.
    1). Show that g.xH=gxH \ (g,x \in G) defines an action of G on X.
    I know that g.xH=gxH \ \Leftrightarrow \ (gx)^{-1} \in G. G is a group closed under the binary operation ".". Therefore if gx \in G, then (gx)^{-1} \in G. This is true so the first criterion for an action is satisfied.

    We next need to show that 1_G.xH=xH \ \forall xH \in X. We know that x \in G and, using the first criterion for an action, we get the required result.

    2). Show that there is a single orbit, X, undder this action.
    By the Orbit-Stabiliser theorem, I know that |G|=|Orb_G(xH)||Stab_G(xH)|.

    I know that Stab_G(xH)=\{g \in G | g.xH=xH \}.

    The elements that satisfy this are going to be all g \in G. Therefore, |Stab_G(xH)|=|G|.

    (^Is this right?)

    This then gives the required result: |G|=|Orb_G(xH)||G| \Rightarrow |Orb_G(xH)|=1.

    3). Show that the stabiliser of the coset xH is xHx^{-1}.
    I need to show that g.xH=xH when g=xHx^{-1}.

    xHx^{-1}xH=xH by substituting in g.

    4). Show that the kernel of the action is a normal subgroup of G that is contained in H.
    I know that \cap_{xH \in X} Stab_G(xH) is the kernel of the action of G on X.

    Rewriting this as \cap_{x \in X} \{g \in G | g.xH=xH \} gives that the only element common to every set is when g=1_G.

    The set \{ 1_G \} is indeed a normal subgroup, and must be in H since every subgroup of a group G shares the same identity element.

    Show that if |G:H|=n with n>1 then G has a quotient group G/N which is isomorphic to a subgroup I of the symmmetric group S_n with |I|>1.
    |G:H|=n \Rightarrow we have n left cosets.

    THerefore choose N to be the subgroup of all left cosets of G. This gives a quotient group G/N.

    Define a homomorphism \phi(g): X \rightarrow X by \phi (g)(xH)=gxH.

    I have a proposition in my notes stating that \phi(g) \in Sym(X) and that \phi:G \rightarrow Sym(X) is a homomorphism.

    From here I get stuck. I've been trying to apply the first isomorphism theorem, but I have no idea how to put it into action.

    Anyway, any help would be appreciated.
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post
    1). Show that defines an action of on .

    I know that g.xH=gxH \ \Leftrightarrow \ (gx)^{-1} \in G. G is a group closed under the binary operation ".". Therefore if gx \in G, then (gx)^{-1} \in G. This is true so the first criterion for an action is satisfied.
    You only need to show (g_1g_2)xH=g_1(g_2xH). You can check that (g_1g_2)xH=g_1(g_2xH), where each g sends each left coset xH to the left coset x'H, and this operation is associative. You might need to pick a representative x from the coset xH and simply say (g_1g_2)x=g_1(g_2x), because if xH=x'H, then x=x'.

    We next need to show that 1_G.xH=xH \ \forall xH \in X. We know that x \in G and, using the first criterion for an action, we get the required result.
    2). Show that there is a single orbit, , undder this action.

    By the Orbit-Stabiliser theorem, I know that |G|=|Orb_G(xH)||Stab_G(xH)|.

    4). Show that the kernel of the action is a normal subgroup of that is contained in .

    I know that Stab_G(xH)=\{g \in G | g.xH=xH \}.

    The elements that satisfy this are going to be all g \in G. Therefore, |Stab_G(xH)|=|G|.

    (^Is this right?)

    This then gives the required result: |G|=|Orb_G(xH)||G| \Rightarrow |Orb_G(xH)|=1.
    No.

    Let M be the set of all left cosets of H in G.
    You need to show that if xH and yH are two arbitrary left cosets in M, then you can find an element g \in G such that gxH=yH. In this case, choose g as yx^{-1}. This means, G acts transitively on M.

    I need to show that g.xH=xH when g=xHx^{-1}.

    xHx^{-1}xH=xH by substituting in g.
    I know that \cap_{xH \in X} Stab_G(xH) is the kernel of the action of G on X.

    Rewriting this as \cap_{x \in X} \{g \in G | g.xH=xH \} gives that the only element common to every set is when g=1_G.

    The set \{ 1_G \} is indeed a normal subgroup, and must be in H since every subgroup of a group G shares the same identity element.
    Let M be the set of all left cosets of H in G. You need to check the induced homomorphism G \rightarrow A(M) defined by g \mapsto \tau_g, where \tau_g:M \rightarrow M and \tau_g(xH)=gxH, where A(M) is the group of permutations of M. If g is in the kernal gxH=xH for all x in G, esp, geH=eH, then g should be in H.



    <br /> <br />
|G:H|=n \Rightarrow we have n left cosets.

    THerefore choose N to be the subgroup of all left cosets of G. This gives a quotient group G/N.

    Define a homomorphism \phi(g): X \rightarrow X by \phi (g)(xH)=gxH.

    I have a proposition in my notes stating that \phi(g) \in Sym(X) and that \phi:G \rightarrow Sym(X) is a homomorphism.

    From here I get stuck. I've been trying to apply the first isomorphism theorem, but I have no idea how to put it into action.

    Anyway, any help would be appreciated.
    The induced homomorphism h: G \rightarrow A(M) defined by the previous question can be used here. The kernel of h is contained in H, actually it has the form N=\bigcap_{x \in G}xHx^{-1} and it is the largest normal subgroup of G contained in H.
    It follows that G/N is isomorphic to the subgroup of A(M), where A(M) is the group of all permutations of M isomorphic to S_n.

    Edit: h is not necessarily a surjective homomorphism. Thus you can't use the first isomorphism theorem here. Rather, you need to show that G/N can be embedded into A(M).
    Last edited by aliceinwonderland; February 18th 2010 at 11:14 PM.
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