Results 1 to 5 of 5

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

  1. #1
    Member
    Joined
    Aug 2008
    Posts
    249

    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 $\displaystyle a \in H $ and $\displaystyle 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 $\displaystyle 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 $\displaystyle a \in H $. but this implies that $\displaystyle a \in H $ and $\displaystyle a \in K $ so $\displaystyle a \in H \cap K $. so the stabilizer of K is the set $\displaystyle 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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    22

    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 $\displaystyle a \in H $ and $\displaystyle 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 $\displaystyle 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 $\displaystyle a \in H $. but this implies that $\displaystyle a \in H $ and $\displaystyle a \in K $ so $\displaystyle a \in H \cap K $. so the stabilizer of K is the set $\displaystyle 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 $\displaystyle HK$. The elements of your orbit are $\displaystyle \{hK:h\in H\}$ which is not equal to $\displaystyle \{hk:h\in H,k\in K\}$ .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 $\displaystyle H\times K$ act on $\displaystyle G$ by $\displaystyle (h,k)\cdot g=hgk^{-1}$. Then, $\displaystyle \mathcal{O}_e=HK$ and $\displaystyle \text{stab}(e)=\{(x,x):x\in H\cap K\}$. Clearly then by the orbit stabilizer theorem $\displaystyle \displaystyle |H||K|=|H\times K|=|HK||H\cap K|$.


    This generalizes to the notion of double cosets if you're interested.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093

    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 $\displaystyle a \in H $ and $\displaystyle 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 $\displaystyle 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 $\displaystyle a \in H $. but this implies that $\displaystyle a \in H $ and $\displaystyle a \in K $ so $\displaystyle a \in H \cap K $. so the stabilizer of K is the set $\displaystyle 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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2012
    From
    hongkong
    Posts
    2

    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??
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,546
    Thanks
    842

    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}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Group actions
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: Oct 8th 2011, 11:29 AM
  2. [SOLVED] Group actions
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Oct 7th 2011, 03:43 PM
  3. [SOLVED] cycle decomposition and group actions.
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: Jan 28th 2011, 02:02 AM
  4. Group actions
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Jul 4th 2009, 04:12 PM
  5. prove A∩B = B∩A
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Aug 17th 2008, 03:28 PM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum