# Math Help - Index/Isomorphism/Quotient Group answer check

1. ## 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.

2. Originally Posted by Showcase_22
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.

$

|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).