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Let $X$ be a topological space on which a group $G$ acts . let $N$ and $K$ be subgroups of $G$. under what condition we have an induced action of $K$ on $X/N$?

My guess: if $N$ is normalized by $K$ in $G$ then we have an induced action. Indeed, we have an action $$K\times X/N\rightarrow X/N;\; (k,[x])\mapsto [kx]$$ This is a well defined action since $[x]$ and $[nx]$ map to the same image $[kx]$ This is because $\forall n\in N$ and $\forall k\in K$ and $\forall x\in X$, $knx=(knk^{-1})kx=n'kx$ where $n'= (knk^{-1})\in N$ since $N$ is normalized by $K$, hence $[knx]=[kx]$.

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1 Answer 1

For simplicity I will assume that $G$ is a discrete group. Then we can just think of a group action of $G$ on $X$ as a homomorphism $\rho : G \to \mathrm{Aut}(X)$, where $\mathrm{Aut}(X)$ is the group of automorphisms of $X$ (in the relevant category). From this point of view, it is obvious that any subgroup of $G$ has an action on $X$.

Now, let $H$ and $K$ be subgroups. In order for $K$ to act on $X / H$, it is necessary that for all $k$ in $K$ and $x$ in $X$, for each $h$ in $H$, $k h \cdot x = h' k \cdot x$ for some $h'$ in $H$. But it is easy to see that this is also sufficient. So $K$ has an induced action on $X / H$ if and only if $\rho(K)$ is contained in the normaliser of $\rho(H)$.

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1) Is your conclusion the same as mine? i mean that we need $K$ is contained in the normalizer of $H$ instead of saying $\rho(K)$ is contained in the normaliser of $\rho(H)$? 2) what does the assumption of $G$ being discrete simplifies here? 3) How do you conclude from the fact that for each $h$ in $H$, $kh.x=h'k.x$ for some $h'$ in $H$. that $\rho(K)$ is contained in the normaliser of $\rho(H)$? –  palio Aug 5 '11 at 7:25
    
@palio: 1. No. Let $G$ act trivially, and let $H$ and $K$ be arbitrary subgroups. 2. If $G$ is not discrete then one is usually required to consider group actions which are continuous as a function on $G$ as well as on $X$. 3. The equation, written without $x$, is just $\rho(kh) = \rho(h' k)$, and the claim follows immediately. –  Zhen Lin Aug 5 '11 at 7:33

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