Take the 2-minute tour ×

I was wondering if we have a finite group $G$ with subgroups $H$ and $K$. With $X:=G/H$. Is there a simple way of finding $X^K$ and when will $X^K=\emptyset$?

Also does this have any baring on when $G/H≅ G/K$?

(This is the first question I've ever asked here so apologies for errors etc.)

share|improve this question
@Peter Brookes: Are $H$ and $K$ normal, or is $X$ the set of left cosets of $H$ in $G$? In the latter case, is the isomorphism in the second paragraph (spelling is "bearing") between $G$-sets, or between quotient groups? –  Arturo Magidin Nov 24 '10 at 4:20
To answer your first question, compute! You can describe $X^K$ quite explicitly by just writing out what it is. –  Mariano Suárez-Alvarez Nov 24 '10 at 4:25
@Arturo Magidin H and K are not necessarily normal, and yes the isomorphism is between quotient groups. –  Peter Brookes Nov 24 '10 at 4:35
@Peter Brookes: if they are not necessarily normal, then you don't have quotient groups, you have sets of left-cosets, and the isomorphism must be an isomorphism of as $G$-sets. If they are quotient groups, then $H$ and $K$ are normal. Which one do you mean? –  Arturo Magidin Nov 24 '10 at 4:47
@Peter, there is absolutely no magic in this, and no need to any technology whatsoever. Alas, Arturo wrote it out for you! What he did was nothing more than using the definition and unfolding it in the concrete situation. You have to learn to do that yourself. –  Mariano Suárez-Alvarez Nov 24 '10 at 5:07

1 Answer 1

up vote 6 down vote accepted

I'm assuming that $G/H$ represents the left cosets of $H$ in $G$, that is the sets of the form $gH$ with $g\in G$, with the action of $G$ on $X$ being given by $x\cdot(gH) = xgH$ (left multiplication), and that $X^K$ is the set of fixed points of $K$; that is, all $gH\in X$ such that for all $k\in K$, $k(gH)=gH$.

If $kgH = gH$, then $g^{-1}kg\in H$. Thus, $gH$ is fixed by all $k\in K$ if and only if $g^{-1}Kg\subseteq H$. So $$X^K = \{ gH\in G/H \mid g^{-1}Kg\subseteq H\}.$$ Note that this is well defined: if $gH = g'H$, then $g'^{-1}g\in H$, so if $g^{-1}Kg\subseteq H$, then $$g'^{-1}Kg' = (g'^{-1}g)(g^{-1}Kg)(g'^{-1}g)^{-1}\subseteq g'^{-1}gH(g'^{-1}g)^{-1}=H.$$

That gives an easy way to find $X^K$ (by looking at the conjugates of $K$) and to figure out if $X^K=\emptyset$ (if no conjugate of $K$ is contained in $H$; e.g., if the order of $K$ is larger than the order of $H$; if $H$ is normal and $K$ is not contained in $H$; many other criteria).

Now, when are the $G$-sets $G/H$ and $G/K$ isomorphic? First, you must have $[G:H]=[G:K]$. Also, there must be a bijection $\psi\colon G/H\to G/K$ such that for all $x,g\in G$, $\psi(gxH) = g\psi(xH)$. In particular, for all $h\in H$ we must have $\psi(H) = \psi(hH) = h\psi(H)$. If $\psi(H) = xK$, then that means that $HxK = xK$, or that $x^{-1}Hx\subseteq K$. In particular, we need a conjugate of $H$ to be contained in $K$. Since $G$ is finite and $H$ and $K$ must have the same order, that means that $H$ is a conjugate of $K$. If $x^{-1}Hx=K$, then for any coset $gH$ we must have $\psi(gH) = g\psi(H) =gxK$.

Conversely, if $H$ is a conjugate of $K$, $x^{-1}Hx = K$, then let $\psi\colon G/H\to G/K$ be the map that sends the coset $gH$ to the coset $gxK$. First, I claim this is well defined: if $yH=zH$, we need to show that $yxK = zxK$. Since $z^{-1}y\in H$, then $x^{-1}z^{-1}yx\in x^{-1}Hx = K$, so $yxK=zxK$, as claimed. If $gH\in G/H$ and $a\in G$, then $\psi(a\cdot gH) = \psi(agH) = agxK = a\cdot gxK = a\psi(gH)$, so $\psi$ is a $G$-set homomorphism. The map $G/K\to G/H$ given by $gK\mapsto gx^{-1}K$ is also a $G$-set homomorphism (as $xKx^{-1}=H$), and is the inverse of the previous map, so $\psi$ is an isomorphism. Thus, the $G$-set $G/H$ is isomorphic to the $G$-set $G/K$ if and only if $H$ and $K$ are conjugate.

Added: It is interesting to note that the definition of $\psi$ above does depend on the choice of $x$; if $x$ and $y$ are two distinct elements such that $xHx^{-1}=yHy^{-1}=K$, then the maps $\psi(gH) = gxK$ and $\phi(gH) = gyK$ are the same map if and only if $y^{-1}x\in K$, that is, if and only if $y$ and $x$ are in the same coset of $K$. So you get one map for each coset of $K$ that intersects the set of elements that conjugate $H$ to $K$.

Note, however, that this does not give you any light into the situation in which $H$ and $K$ are normal, and you are interested in knowing whether the quotient groups $G/H$ and $G/K$ are isomorphic as groups: obviously, if $H$ and $K$ are normal, then they are conjugate if and only if they are equal, but it is trivial to find examples of two distinct normal sugroups of a finite group $G$ that give you isomorphic quotients. For example, take any two distinct nontrivial proper subgrups of the Klein $4$-group.

share|improve this answer
In your displayed equation, you meant $gH\in G/H$ inside the set. –  Mariano Suárez-Alvarez Nov 24 '10 at 5:07
@Mariano: Thanks. –  Arturo Magidin Nov 24 '10 at 5:09

Your Answer


