# 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 $H$

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!

• $G/H$ normally denotes the set of left cosets $xH$ not right cosets. Feb 16 at 6:57
• I'll fix that - I intend to work with right cosets. Is $H\backslash G$ the correct notation, or is it $G\backslash H$? Feb 16 at 6:59
• Right cosets are $Hx$, so you need the $H$ on the left and the $G$ on the right, as in $H\backslash G$. The map you want is $f_g(Hx) = Hxg^{-1}$. Feb 16 at 8:31
• Thanks! I'll keep that in mind. Feb 16 at 9:16

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 $$\phi_{gh} = \phi_g \circ \phi_h$$.

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

• Could you please explain why: "...if $Hxg^{-1} = Hyg^{-1}$, then$^\color{blue}1$ $xg^{-1}(yg^{-1})^{-1} = xy^{-1}$ is in $H$..." - I had messed this part up in my attempt. Feb 16 at 8:28
• The explanation for that part is in the footnote. Just put $N=H$, $a = xg^{-1}$ and $b = yg^{-1}$. Also note that $(yg^{-1})^{-1} = gy^{-1}$. Feb 16 at 8:38
• Thanks, that's helpful! Any thoughts on the second part, about $\ker\phi$? I know that $\ker\phi = \{g\in G: f_g = I\}$, i.e. we want $f_g(Hx) = Hxg^{-1} = Hx$ for all $Hx \in S$. Feb 16 at 9:17
• Note that \begin{align} \ker \phi &= \{g \in G : (\forall x \in G) \ Hxg^{-1} = Hx\} \\ &= \{g \in G : (\forall x \in G) \ xg^{-1}x^{-1} \in H\} \\ &= \{g \in G : (\forall x \in G)(\exists h \in H) \ xg^{-1}x^{-1} = h\} \\ &= \{g \in G : (\forall x \in G)(\exists h \in H) \ g = x^{-1}h^{-1}x\} \\ &= \{g \in G : (\forall x \in G) \ g \in x^{-1}Hx\} \\ &= \bigcap_{x \in G} x^{-1}Hx (\subseteq e^{-1}He = H). \end{align} Does this help? (I corrected some equalities). Feb 16 at 9:38
• @strawberry-sunshine And that completes the proof, well done! Feb 16 at 18:31