If $H\le G$, there is a homomorphism $\phi:G\to A(H\backslash G)$ and $\ker\phi$ is the largest normal subgroup of $G$ contained in $H$

Question

Let $G$ be a group. If $H\le G$ (notation for $H$ is a subgroup of $G$), then show that there is a homomorphism $\phi: G\to A(H\backslash G)$ and that $\ker\phi$ is the largest normal subgroup of $G$ contained in $H$. Here, $A(H\backslash G)$ denotes the set of all permutations on the set $H\backslash G$, of all right cosets of $H$ in $G$.

For simplicity, let $S := H\backslash G$ hereafter. We know that $S = \{Hg: g\in G\}$ where $Hg:= \{hg:h\in H\}$. Consider the following mapping which I claim is a homomorphism: $$\phi: G\to A(S), \phi: g\mapsto f_g$$ where $$f_g:S\to S, f_g: Hx\mapsto Hxg$$ $f_g$ is a bijection (permutation on $S$) because:

1. Injective. $f_g(Hx) = f_g(Hy) \implies Hxg = Hyg$. This happens iff $xg(yg)^{-1}\in H$ which gives $xy^{-1}\in H$ and thus $Hx=Hy$.
2. Surjective. Consider a right coset $Hx$. Clearly, $f_g(Hxg^{-1}) = Hxgg^{-1} = Hx$ showing surjectivity. Note that $xg^{-1}\in G$ since $G$ is a group, so $Hxg^{-1}$ is a valid right coset to choose.

I'm stuck while showing $\phi$ is a homomorphism because:

1. Take $g,h\in G$. We want $\phi(gh) = f_{gh} = f_g\circ f_h = \phi(g)\phi(h)$, but: consider arbitrary $Hx\in G/H$. $f_{gh}(Hx) = Hxgh = (Hxg)h = f_h(Hxg) = f_h(f_g(Hx))$ which is $f_h\circ f_g$ and not what we need. Is this not the right homomorphism? What do I do?

2. Certainly $\phi(g)$ is inside $A(S)$, so that is not a problem.

Lastly, I don't know what $\phi$ is yet (though intuition says it's closely related to the above stuff), leave alone $\ker\phi$. In the second part of the theorem, we want to show that $\ker\phi$ is the largest normal subgroup, i.e. any other normal subgroup of $G$ in $H$ is surely contained in $\ker\phi$. I hope this is easy to show once $\phi$ is known!

I'd appreciate any help, thank you!

Answer 1

Let $g \in G$, and consider the mapping $\phi_g : S \to S$ given by $Hx \mapsto Hxg^{-1}$.

• $\phi_g$ is injective, for if $Hxg^{-1} = Hyg^{-1}$, then$^\color{blue}1$ $xg^{-1}(yg^{-1})^{-1} = xy^{-1}$ is in $H$, but this means precisely that $Hx = Hy$.
• $\phi_g$ is surjective, since $Hx = \phi_g(Hxg)$ for all $x \in G$.

Thus $\phi_g \in A(S)$ for each $g \in G$, and then the map $\phi : G \to A(S)$ given by $g \mapsto \phi_g$ is well-defined. Finally, $\phi$ is a group homomorphism: since $$(\forall x \in G) \quad \phi_{gh}(Hx) = Hx(gh)^{-1} = Hxh^{-1}g^{-1} = \phi_g(Hxh^{-1}) = \phi_g(\phi_h(Hx))$$ we have $\phi_{gh} = \phi_g \circ \phi_h$.

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$^\color{blue}1$ Recall that $Na = Nb$ iff $ab^{-1} \in N$.