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Help with a group theory proof

  1. Aug 20, 2005 #1
    I need some help with what should be a really simple group theory proof, but for some reason I'm hitting my head against a wall on this one. I seem to be missing something simple to get me to the next step.

    I'm learning this stuff in Swedish, so I'm not sure about all the English words, but I'll try and make it at least clear. The problem is pretty simple. We just learned the Lagrange theorem and a few simple resulting theorems from that.

    Let H and K be two normal cyclical subgroups to a finite group G. Let |H| and |K| be relatively prime. Show that:

    HK = {hk | h element of H, k element of K}

    is also a cyclical subgroup to G.

    Showing it's a subgroup doesn't seem to be much problem, but showing it is cyclical is tripping me up. I _think_ I can do it if I show hk=kh for all h and k, but I'm not sure how to do this. I'm also not sure what the groups being normal has to do with it (if anything) since aren't all cyclical groups automatically normal (since they always commute, and all groups that commute are normal)?

    I can see intuitively that the product of the generators to H and K should itself be a generator to HK, but I'm having trouble "proving" it more formally.

    This feels like it should be really simple, with one little step I'm missing, but I can't see it.

    Things I played around with: H and K have no undergroups in common, and I believe they have to be disjoint (one follows from the other I think, doesn't it?). I can see that the only hk that equals 1 is when they both equal one, which means that if hk commutes then (hk)^|H||K| = 1.

    Anyway, any help or clues would be appreicated. Thanks.
  2. jcsd
  3. Aug 20, 2005 #2

    matt grime

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    This is a false assertion. The commutativity of a subgroup, which is purely an internal property, has nothing to do with it being normal, which is a global property with respect to the other elements. C_2 in S_3 is not normal for instance (and shows a counter example exists for the smallest possible orders, incidentally)

    if H and K are cyclic and normal we aim to show that HK is cyclic (not necessarily normal), so try to find a generator, if h generates H and k generates K (with orders p and q ersp) can you think of any other element that hk that will generate HK? No, nothing springs to mind so try finding its order, or showing that the elements (hk)^r are distinct for 1<=r<=pq
  4. Aug 20, 2005 #3

    matt grime

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    subgroups, and they always have the trivial group in their intersection. and apart from this they may or may not have elements in common.

    not disjoint, no, subgroups are never disjoint, see above.
  5. Aug 20, 2005 #4
    sorry, I meant besides the identity element in the other comments (about disjoint for example). or the trivial group as I think you called it.

    I think I got confused with the commuting ... if the larger group commutes, then all subgroups must be normal, right? This is directly from a therom in my book. But maybe it's irrelevant anyway.

    The problem I'm having with showing that all (hk)^r are distinct is that I really want hk to commute so that (hk)^r = h^r * k^r, in which case I'm pretty sure I can do this. But without showing that they commute, I'm having a hard time showing that all the elements have to be distinct. The order of <hk> would have to be pq, but again that would be easier to show if hk commuted again.

    I think I can show that for h and k not equal to 1, that hk can't be an element of H or K (talking general h and k here, not the generators), though I'm not sure if this helps.

    of course this leads me to thinking that since all h and all k are also elements of HK, that h^n * k^m has to be also for all integers n and m.

    I'm still missing something to put it all together though.
  6. Aug 20, 2005 #5

    matt grime

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    h an dk certainly do not necessarily commute. nor can you conclude the hk is not in H or K. indeed it may happen (My mistake; wrote too quickly) that HK does not have order pq (eg if H is a subgroup of K).
  7. Aug 20, 2005 #6
    But they are relatively prime, so one can't be a subgroup of the other (ignoring the trivial group again), right? Subgroup length has to be a divisor of group length.

    Fine, h and k don't commute (although if HK is cyclical they will anyway, but sure, after the fact). How do I do this otherwise?
  8. Aug 20, 2005 #7
    never mind on the commuting, I realize that it just means one hk times another hk comute in HK, not that h and k themselves commute.

    Doesn't matter though, I still need positive help on solving this.
  9. Aug 20, 2005 #8
    I was also trying to think about why they mentioned the groups were normal. The only theorem for normal groups that seems like it might be helpful is that

    ghg^-1 is an element of H for all g element of G and h element of H. I thought i could get somwhere using elements of K for g, but it dead ended for me.

    There must be just some simple thing I'm missing here.
  10. Aug 20, 2005 #9


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    Well, it also tells you HK = KH, doesn't it?
  11. Aug 20, 2005 #10


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    Here's a hint:

    1) H and K are normal, so for any h in H (in fact, any g in G would do), and for any k in K, hkh-1 is in K.
    2) Given that |H| and |K| are relatively prime, what can you conclude about the intersection of H and K?
    3) hkh-1k-1 = h(kh-1k-1) = (hkh-1)k-1.
  12. Aug 21, 2005 #11

    1) right, I was playing with that from the start,but nothing jumped out at me
    2) I mentioned that above, their intersection is just {1}, right?
    3) this confused me, this seemed to say that the product of any two elements of H is equal to the product of two members of K, which doesn't seem like it can be right.
    h(kh-1k-1) is just h times some element of H, and (hkh-1)k-1 is just the inverse of k (which is just some member of K) times some other member of K.


    since that doesn't mean that each individual h and k commute, I don't see how that helps me? What am I missing?
  13. Aug 21, 2005 #12


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    Well, it tells you that you can write any kh in the form h'k'. I was thinking that might be a useful fact. (Though, I can't say that I've worked through this proof any time in recent history, so I can't be positive)

    Think some more about what you've just said...
  14. Aug 21, 2005 #13

    matt grime

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    apologies, i missed the |H| ans |K| but relatively prime. must've not been thinking that clearly since this is a problem that i actually once set on some sheet of group theory questions.
  15. Aug 21, 2005 #14
    But that seems impossible to me if the intersection of H and K is {1}. Oh, wait, that must mean those quantities are both equal to 1 (the only way a product of two elements of H can equal the product of two elements of K), which means:


    which means

    hk=kh, and they commute!
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