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!

  • $\begingroup$ $G/H$ normally denotes the set of left cosets $xH$ not right cosets. $\endgroup$
    – Derek Holt
    Feb 16 at 6:57
  • $\begingroup$ I'll fix that - I intend to work with right cosets. Is $H\backslash G$ the correct notation, or is it $G\backslash H$? $\endgroup$ Feb 16 at 6:59
  • $\begingroup$ 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 $f_g(Hx) = Hxg^{-1}$. $\endgroup$
    – Derek Holt
    Feb 16 at 8:31
  • $\begingroup$ Thanks! I'll keep that in mind. $\endgroup$ 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 $ab^{-1} \in N$.

  • $\begingroup$ 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. $\endgroup$ Feb 16 at 8:28
  • $\begingroup$ The explanation for that part is in the footnote. Just put $N=H$, $a = xg^{-1}$ and $b = yg^{-1}$. Also note that $(yg^{-1})^{-1} = gy^{-1}$. $\endgroup$
    – azif00
    Feb 16 at 8:38
  • $\begingroup$ 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 $Hx \in S$. $\endgroup$ Feb 16 at 9:17
  • $\begingroup$ 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). $\endgroup$
    – azif00
    Feb 16 at 9:38
  • 1
    $\begingroup$ @strawberry-sunshine And that completes the proof, well done! $\endgroup$
    – azif00
    Feb 16 at 18:31

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