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Math Help - prove that |HK| = |H||K| / |H ∩ K| (using group actions)

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    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.
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    MHF Contributor Drexel28's Avatar
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    Re: prove that |HK| = |H||K| / |H ∩ K| (using group actions)

    Quote Originally Posted by oblixps View Post
    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.
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    MHF Contributor Amer's Avatar
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    Re: prove that |HK| = |H||K| / |H ∩ K| (using group actions)

    Quote Originally Posted by oblixps View Post
    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
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    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??
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    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}.
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