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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 $\displaystyle G$.

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.Quote:

1). Show that $\displaystyle g.xH=gxH \ (g,x \in G)$ defines an action of $\displaystyle G$ on $\displaystyle X$.

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.

By the Orbit-Stabiliser theorem, I know that $\displaystyle |G|=|Orb_G(xH)||Stab_G(xH)|$.Quote:

2). Show that there is a single orbit, $\displaystyle X$, undder this action.

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: $\displaystyle |G|=|Orb_G(xH)||G| \Rightarrow |Orb_G(xH)|=1$.

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

3). Show that the stabiliser of the coset $\displaystyle xH$ is $\displaystyle xHx^{-1}$.

$\displaystyle xHx^{-1}xH=xH$ 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 $\displaystyle X$.Quote:

4). Show that the kernel of the action is a normal subgroup of $\displaystyle G$ that is contained in $\displaystyle H$.

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.

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

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 $\displaystyle |I|>1$.

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.