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

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

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

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

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

    (^Is this right?)

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

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

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

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

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

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

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

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

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

    I have a proposition in my notes stating that $\displaystyle \phi(g) \in Sym(X)$ and that $\displaystyle \phi:G \rightarrow Sym(X)$ 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$\displaystyle g.xH=gxH \ \Leftrightarrow \ (gx)^{-1} \in G$. $\displaystyle G$ is a group closed under the binary operation ".". Therefore if $\displaystyle gx \in G$, then $\displaystyle (gx)^{-1} \in G$. This is true so the first criterion for an action is satisfied.
    You only need to show $\displaystyle (g_1g_2)xH=g_1(g_2xH)$. You can check that $\displaystyle (g_1g_2)xH=g_1(g_2xH)$, 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 $\displaystyle (g_1g_2)x=g_1(g_2x)$, because if $\displaystyle xH=x'H$, then $\displaystyle x=x'$.

    We next need to show that $\displaystyle 1_G.xH=xH \ \forall xH \in X$. We know that $\displaystyle x \in G$ 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 $\displaystyle |G|=|Orb_G(xH)||Stab_G(xH)|$.

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

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

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

    (^Is this right?)

    This then gives the required result: $\displaystyle |G|=|Orb_G(xH)||G| \Rightarrow |Orb_G(xH)|=1$.
    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 $\displaystyle g \in G$ such that $\displaystyle gxH=yH$. In this case, choose g as $\displaystyle yx^{-1}$. This means, G acts transitively on M.

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

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

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

    The set $\displaystyle \{ 1_G \}$ is indeed a normal subgroup, and must be in $\displaystyle H$ since every subgroup of a group $\displaystyle G$ 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 $\displaystyle G \rightarrow A(M)$ defined by $\displaystyle g \mapsto \tau_g$, where $\displaystyle \tau_g:M \rightarrow M$ and $\displaystyle \tau_g(xH)=gxH$, 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.



    $\displaystyle

    |G:H|=n \Rightarrow$ we have $\displaystyle n$ left cosets.

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

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

    I have a proposition in my notes stating that $\displaystyle \phi(g) \in Sym(X)$ and that $\displaystyle \phi:G \rightarrow Sym(X)$ 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 $\displaystyle h: G \rightarrow A(M)$ defined by the previous question can be used here. The kernel of h is contained in H, actually it has the form $\displaystyle N=\bigcap_{x \in G}xHx^{-1}$ 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 $\displaystyle S_n$.

    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; Feb 18th 2010 at 10:14 PM.
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