Thread: prove that |HK| = |H||K| / |H ∩ K| (using group actions)

1. prove that |HK| = |H||K| / |H ∩ K| (using group actions)

I want to prove that |HK| = |H||K| / |H ∩ K| using group actions. First, I let H and K be subgroups of G and the set S be defined as the set of left cosets of K in G. I define a group action by a * (xK) = axK where $a \in H$ and $x \in G$.

by considering the orbit of the element K in S, i have that the orbit of K is the set of all s = a*K = aK, such that $a \in H$. but this is just the set theoretic product HK, so the orbit of K is equal to HK. From the orbit stabilizer theorem, the number of elements in the orbit of K is equal to the number of left cosets of the stabilizer of K. The stabilizer of K is the set of all a such that a*K = aK = K and $a \in H$. but this implies that $a \in H$ and $a \in K$ so $a \in H \cap K$. so the stabilizer of K is the set $H \cap K$.

now the number of left cosets of the stabilizer of K is |H|/|H ∩ K|. therefore the orbit stabilizer theorem says that |HK| = |H|/|H ∩ K|, but i seem to be missing a factor of |K| somewhere and going through the problem I can't seem to find where it must come from. Can someone help show me what went wrong? thanks.

2. Re: prove that |HK| = |H||K| / |H ∩ K| (using group actions)

Originally Posted by oblixps
I want to prove that |HK| = |H||K| / |H ∩ K| using group actions. First, I let H and K be subgroups of G and the set S be defined as the set of left cosets of K in G. I define a group action by a * (xK) = axK where $a \in H$ and $x \in G$.

by considering the orbit of the element K in S, i have that the orbit of K is the set of all s = a*K = aK, such that $a \in H$. but this is just the set theoretic product HK, so the orbit of K is equal to HK. From the orbit stabilizer theorem, the number of elements in the orbit of K is equal to the number of left cosets of the stabilizer of K. The stabilizer of K is the set of all a such that a*K = aK = K and $a \in H$. but this implies that $a \in H$ and $a \in K$ so $a \in H \cap K$. so the stabilizer of K is the set $H \cap K$.

now the number of left cosets of the stabilizer of K is |H|/|H ∩ K|. therefore the orbit stabilizer theorem says that |HK| = |H|/|H ∩ K|, but i seem to be missing a factor of |K| somewhere and going through the problem I can't seem to find where it must come from. Can someone help show me what went wrong? thanks.
The problem is that your orbit is not $HK$. The elements of your orbit are $\{hK:h\in H\}$ which is not equal to $\{hk:h\in H,k\in K\}$ .One's a set of cosets one is a set of group elements(don't worry, I had to stare at it for five minutes too. I was like "this guy just disproved math". haha, we all do stupid things, no worries).

I would suggest letting $H\times K$ act on $G$ by $(h,k)\cdot g=hgk^{-1}$. Then, $\mathcal{O}_e=HK$ and $\text{stab}(e)=\{(x,x):x\in H\cap K\}$. Clearly then by the orbit stabilizer theorem $\displaystyle |H||K|=|H\times K|=|HK||H\cap K|$.

This generalizes to the notion of double cosets if you're interested.

3. Re: prove that |HK| = |H||K| / |H ∩ K| (using group actions)

Originally Posted by oblixps
I want to prove that |HK| = |H||K| / |H ∩ K| using group actions. First, I let H and K be subgroups of G and the set S be defined as the set of left cosets of K in G. I define a group action by a * (xK) = axK where $a \in H$ and $x \in G$.

by considering the orbit of the element K in S, i have that the orbit of K is the set of all s = a*K = aK, such that $a \in H$. but this is just the set theoretic product HK, so the orbit of K is equal to HK. From the orbit stabilizer theorem, the number of elements in the orbit of K is equal to the number of left cosets of the stabilizer of K. The stabilizer of K is the set of all a such that a*K = aK = K and $a \in H$. but this implies that $a \in H$ and $a \in K$ so $a \in H \cap K$. so the stabilizer of K is the set $H \cap K$.

now the number of left cosets of the stabilizer of K is |H|/|H ∩ K|. therefore the orbit stabilizer theorem says that |HK| = |H|/|H ∩ K|, but i seem to be missing a factor of |K| somewhere and going through the problem I can't seem to find where it must come from. Can someone help show me what went wrong? thanks.
K is not an a single element it is a set which has l K l elements
you can check this thread look into post #2

4. Re: prove that |HK| = |H||K| / |H ∩ K| (using group actions)

I want to ask do HK is a subgroup of G ??
If yes, DO we need to assume one of the H, K is normal subgroup of G to ensure HK is normal in G ??
If HK is not a subgroup of G, will the equality become Inequality??

5. Re: prove that |HK| = |H||K| / |H ∩ K| (using group actions)

HK is usually NOT a subgroup of G. if one of H,K is normal in G, then HK will be a subgroup:

let's say K is normal:

then (hk)(h'k') = h(kh')k', and since K is normal, Kh' = h'K, so kh' = k"h', for some k" in K, so

h(kh')k' = h(h'k")k' = (hh')(k"k'), which is in HK, so HK is closed under multiplication.

note that (hk)-1 = k-1h-1.

since Kh-1 = h-1K, k-1h-1 = h-1k', for some k' in K, so:

(hk)-1 = h-1k', which is in HK (since h-1 is in H, because H is a subgroup).

thus HK contains all inverses of elements hk in HK, and is thus a subgroup of G.

IF HK is a subgroup, there is still no guarantee that HK will be a NORMAL subgroup.

for example H = {e, (1 2)(3 4)} and K = {e, (1 3)(2 4)} are both subgroups of A5, and HK = {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}

is indeed a subgroup of A5, but A5 is simple as has no normal subgroups.

indeed, even the normality of H (or K) does not guarantee the normality of HK,

in D6, the dihedral group of order 12, we have that <r3> is normal in D6,

so if we take H = <r3> = {1, r3} and K = {1,s}, w have that HK = {1, r3, s, r3s} is a subgroup of D6.

however, this subgroup is not normal, consider the left coset r(HK) =

{r,r4, rs, r4s}, while the right coset (HK)r =

{r, r4, r5s, r2s}, so r(HK) and (HK)r are different, thus HK is not normal.

i'm not sure what you mean about "equality", as in the equation:

|HK| = |H||K|/|H∩K|, there is no "inequality". and the normality of H and K has nothing to do with the size of H∩K:

let G = Z12, the group of integers mod 12. this group is abelian, so EVERY subgroup is normal.

let H = <2> = {0,2,4,6,8,10}, and let K = <3> = {0,3,6,9}.

even though both groups are normal, we have a non-trivial intersection: H∩K = {0,6}.