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Prove that a group of order (m^2)(n^2) is abelian if it has all normal subgroups

  1. Dec 7, 2009 #1
    1. The problem statement, all variables and given/known data
    Let G be a group with order (m^2)(n^2), where m and n are primes and m not equal to n. Prove that every subgroup of G is normal if and only if G is abelian.

    3. The attempt at a solution

    The <== direction is simple (I think), since all subgroups of an abelian group are normal.
    The ==> direction I can't seem to figure out.

    Using Sylow's theorems, I determined that there are subgroups of G of order m, m^2, n, and n^2, and that the group of order m is normal in the group of order m^2 and the group of order n is normal in the group of order n^2. However, I haven't gleaned anything useful from Sylow.

    Next I tried supposing that all the subgroups of G were normal and that, contrarily, there existed a, b in G such that ab != ba, and then I played around with different conjugations. Here were two things in particular I tried:
    1) Pick 2 arbitrary elements in a normal group H and conjugate them by x and y respectively.
    2) Pick a in a normal subgroup A and b in a normal subgroup B such that a, b are not equal. Then conjugate a and b by arbitrary elements of G.

    What I am trying to do is derive a contradiction - either that the a and b I chose must commute, or that there is a subgroup in G which is not normal. I have not had any luck so far.

    Some crazier things I tried involved the commutator subgroup, which must be nontrivial. I thought possibly I could create a homomorphism f from some group G* to G with the commutator subgroup of G* being a subgroup of ker(f). This would imply that Im(f) = G is an abelian group. I think this path is too complicated though.

    Any suggestions?

  2. jcsd
  3. Dec 7, 2009 #2
    as you said, the first direction is very easy...and you said it correctly...
    about the second direction:
    if you have a subgroup of a prime order, can you say anything about it's content? you know from lagrange's theorem, that each and every one of it contents has order p or order 1. but p is larger then 1 so there is an a in the group that is of order p.
    The order of a is equal to the order of the group . ( o(a)=o(G) ) ... So the group is cyclic....and what do you know about cyclic sub-groups? :)

    HOPE I helped....
  4. Dec 7, 2009 #3

    Are they normal? I know that all subgroups of a cyclic group are cyclic.
    I already know from Sylow's theorems that the group of order m is normal in the group of order m^2 and the group of order n is normal in the group of order n^2.
    The groups of order m and order n are the only groups with prime order. I need to show that the groups of order m^2 and n^2 are normal in G, but they aren't also cyclic, are they?
    By Sylow's 2nd theorem, any two Sylow p-subgroups are conjugate, but I don't know if I can say more without knowing the relationship between m and n.

    Am I just looking in the wrong direction?
  5. Dec 8, 2009 #4
    I think you're just a bit confused...
    You've handled one direction correctly...we need to prove now that if every sub-group of G is normal then G is abelian...
    By the facts I've told you- you know that each and every one of the sub-groups fro a prime order are cyclic...Cyclic groups are also abelien (you can prove it easily), so these sub-groups are abelian...take one sub-group and an element from a different sub group. you get: aH=Ha... you will finally get that the whole group G must be abelian...
  6. Dec 8, 2009 #5
    Thanks for your help. Yes, I am confused. What you're saying makes sense. I think I almost have it, but there are two things which I'm still not sure about.

    1) It is true that if a subgroup H is abelian (i.e. the elements of H commute with one another), then the elements of H also commute with the elements of G? I don't think this is true, but are you saying that this isn't even necessary?

    2) Yes, all cyclic groups and subgroups are abelian, but are all p-groups and p-subgroups cyclic? In my group I will have a subgroup of order m^2, which will also have a subgroup of m.

    Would this work?
    Let H be a subgroup of G. Then |H| is either m^k or n^k, for some integer k, so H is cyclic, and hence abelian.
    Let a be in G and h be in H. Then since H is abelian, ah = ha. Therefore aH = Ha for all a in G and all subgroups H of G.
  7. Dec 8, 2009 #6
    1 isn't true ofcourse...

    about 2... take a p-group of order p... from lagrange theorem we know that each and every one of it's elements have an order that divides p... so their order is 1 or p...hence there's an element in that p group that has order p...That's excatly the definition of a cyclic group!!! You have an element that has the same order as the group! You can prove by eas. that this element is the generator of the group...
    So you get that all the p-sub-groups are cyclic -> abelian...

    OK so far?
  8. Dec 8, 2009 #7
    I see that if you have a p-subgroup of order p, then it is cyclic, but what if you have a p-subgroup of order p*p? Then there are some elements of order p and an element of order p*p. Can you still show that the element of order p*p is the generator of the group and that the p-subgroup of order p*p is cyclic? If not then you can't state that all p-subgroups are cyclic.

    Why am I not understanding this?
  9. Dec 8, 2009 #8
    You understand everything correctly...I was wrong in the way I wrote my last message.
    When I wrote p-sub-group, I meant all the groups from order p (I know it isn't the definition, but I used very intuitive concepts...)...
    So...we know this far, that each and every one of the sub-groups that has a prime order are cyclic. You know that each and every one of the subgroups is normal...
    Please write on a paper the definition of a normal group and use the fact that the prime-ordered-subgroups are abelian...I am pretty sure you'll see write away the point... "normality" gives you aH=Ha for every a in G.... that means that the abelian sub-groups commute with each other!!!

    BTW-subgroups of order p^2 aren't necessarily cyclic (as you said...)

    Hope you'll get it this time... but if you don't - dont hesitate and write here again...I'll keep "aiming" you to the soloution 'till will get the point...
  10. Dec 9, 2009 #9
    The normality of each subgroup gives you aH = Ha for every a in G, which means that the subgroups commute as subgroups, but the individual elements do not necessarily commute. For some element n of a normal subgroup N and an element g in G, gn = gn2 for some n2 in N, but n is not necessarily equal to n2.

    Suppose you had a subgroup H, and some elements a and b in G. Then (ab)H=(aH)(bH)=(Ha)(Hb)=H(ab), and (ab)H = (ba)H, since H is closed. However, this doesn't imply that the individual elements commute.

    I don't see how to get from aH = Ha to ah = ha, for all a in G.
  11. Dec 9, 2009 #10
    The groups themselves are abelian (I proved they're cyclic...) and every element in the subgroup commute with every element outside the subgroup...It's the exact definition for commutativity ...
  12. Dec 9, 2009 #11
    I understand the elements in the subgroup commute amongst themselves, but saying aH = Ha does not show that they commute with the other elements. You could have a * h1 = h2 * a, for different h1, h2 in H. How do you use this to show that each individual element of H commutes?
  13. Dec 9, 2009 #12
    Do you know the definition of multiplicating an element with a group?
    For an element a in G, and a sub-group H of G : aH={ah | h is in H} ...
  14. Dec 9, 2009 #13
    Yes I know that definition. H is a collection of elements, and aH is the collection of all those elements left-multiplied by a. But you can't do things like: aH = bH implies a = b, and you can't say that if aH = Ha, then ah = ha, for all h in H. You can say that for h1 in H, h1 left-multiplied by some element a is equal to some (possibly different) element h2 of H right-multiplied by a.
  15. Dec 10, 2009 #14
    Hey there... Don't be angry at me please...I'm only trying to help you...
    I didn't say anything such as aH=bH implies a=b... Maybe I was wrong...
    Wait until someone else will help you...

    Sry I couldn't help more...
  16. Sep 7, 2010 #15
    Okay, I really don't know what TheForumLord is on about, but I don't think her/his approach is going to work.

    Sprooth, you had a really good idea when you first started. As you said, there exist subgroups M and N of order M² and N², respectively. Now, since m and n are distinct primes, the intersection of M and N is trivial. Now, there is a theorem that states that the product of two normal subgroups with trivial intersection is the internal direct product of those groups, so G≃M×N. Now, since M and N are both abelian (why?), this implies that G is abelian. Q.E.D.
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