2
$\begingroup$

So I have this example in my thesis that you can define an action of a group $G$ on G/H (where $H \lhd G$) like this: $\phi_g(aH)=gaH$. I thought it was simple enough that I didn't need to prove it's true, except now I tried and I can't prove it!

We say that $G$ acts on $G/H$ if there's a homomorphism $\phi: G:\to Aut G/H$, right? Well, so I need to prove two things: $\phi$ is a homomorphism from $G$ to $Aut G/H$, and $\phi_g$ is an automorphism of $G/H$. I can prove the first claim, but not the second.

If $\phi_g$ is an automorphism of $G/H$, then $$\phi_g(aHbH)=\phi_g(aH)\phi_g(bH),$$

but $\phi_g(aHbH)=\phi_g(abH)=gabH$, while $\phi_g(aH)\phi_g(bH)=gaHgbH=gagbH$, and I don't see why $gabH$ should be the same as $gagbH$. I'm clearly missing something really simple, but I really don't know what!

$\endgroup$
  • 1
    $\begingroup$ $\phi$ does not have to be a morphism to $Aut(G/H)$ but to $Sym(G/H)$, since you have an action of $G$ on the set $G/H$. Moreover what should $Aut(G/H)$ be? In the same vein, you write $aHbH = abH$ but this is only true for a normal subgroup $H$. $\endgroup$ – M.U. Oct 5 '16 at 8:47
  • 1
    $\begingroup$ Maybe a bit easier to work with is this (equivalent) definition: A (left) action of a group $G$ on a set $X$ is given by a map $G\times X\rightarrow X$, $(g,x)\mapsto g\cdot x$ satisfying: (1) $1\cdot x = x$ for all $x\in X$ and (2) $g\cdot (h\cdot x) = (gh)\cdot x$ for all $g,h\in G$, $x\in X$. $\endgroup$ – Claudius Oct 5 '16 at 8:56
  • 1
    $\begingroup$ @Nicola. No you don't. You have a group $G$ acting on a set $X$ (in your case $G/H$ but it doesn't matter). Having a group action (e.g. like user218931 wrote) is the same thing as having a morphism from $G$ to $Sym(X)$, i.e. the set of bijections $X \to X$. Do you see the important difference? It seems to me as you may again work through the definition of a group action as you may have not unterstood correctly. $\endgroup$ – M.U. Oct 5 '16 at 9:08
  • 1
    $\begingroup$ $G$ is acting on $G/H$ viewed as a set, not as a group (even if $H$ is normal in $G$, so that $G/H$ is a group)! In this case, the definition of Suzuki doesn't apply. Please refer to the comments of @M.U. $\endgroup$ – Claudius Oct 5 '16 at 9:50
  • 2
    $\begingroup$ If you really want to make $G$ act on $G/H$ as a group ($H$ normal in $G$), then you might check $(g,aH)\mapsto gag^{-1}H$. $\endgroup$ – Claudius Oct 5 '16 at 9:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.