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Subgroup of a Direct Product

  1. Mar 4, 2008 #1
    1. The problem statement, all variables and given/known data
    Let M and N be normal subgroups of G, and suppose that the identity is the only element in both M and N. Prove that G is isomorphic to a subgroup of the product [itex]G/M\times G/N[/itex]

    2. Relevant equations
    Up until now, we've dealt with isomorphism, homomorphisms, automorphisms, Lagrange's Theorem, and other bits and pieces of theorems.

    3. The attempt at a solution

    I have no idea how to start this! I've looked over my notes and couldn't find anything obvious. I know that |M| divides |G| and the same goes for |N|. Does the fact that M and N only have {e} as a common element mean that |N| and |M| are relatively prime?

    Also, can anybody explain exactly what, say, G/M is doing? My notes mention that it means "the cosets of M in G", but I'm not sure how to deal with it. Is there an official name for it so I can look up other information on it? Thanks!
  2. jcsd
  3. Mar 4, 2008 #2


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    G/M is called a "quotient group". It's the group you get once you basically set M to zero.

    If M and N intersect in only {e}, then that doesn't mean |M| and |N| are coprime. For example in the group C_2 x C_2 = {e, a, b, c} (the Klein four group), {e, a} and {e, b} are subgroups of order 2 that intersect only in {e}.

    How about you try to construct an injective homomorphism into G/M x G/N?
  4. Mar 4, 2008 #3
    My understanding of quotient groups is still fuzzy... we don't have a textbook other than a supplemental online textbook, and from what I've gathered from it, G/N is only a quotient group if N is normal in G (is this right?). So what if N is not normal in G... what exactly is this object and how can I use it? I think if I understand this, it may help...

    I can also see that the product MN is an abelian subgroup of G. My instructor also stated (without proof) the following two properties:

    1) Every finite abelian group is a direct product of cyclic groups of prime power order
    2) The prime powers that occur are uniquely determined, up to reordering

    Can I use these two facts at all (especially since MN is an abelian subgroup of G)?

    As another idea, I tried going back over my notes and came up with something else of interest. We showed in class that for a homomorphism [itex]\phi:G\rightarrow H[/itex], the image of G is a subgroup of H. If the homomorphism if injective, then it is certainly surjective onto its own image, which means that it is an isomorphism. Is this what you were suggesting, morphism?

    If that's the case, then all I need are surjective homomorphisms from G to G/M and G/N, right? But I can't really do that until I understand quotient groups a bit more... :( I'll keep at it! Thanks for your help morphism!

    EDIT: Whoops, I meant I need injective homomorphisms. Sorry!
    Last edited: Mar 4, 2008
  5. Mar 4, 2008 #4


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    Let's introduce an equivalence relation on G. We'll say that g~h iff gh-1 is in N. This is motivated by our desire to make N=e. Consider the set G/~ of equivalence classes of G under ~. Let's use the notation gN to denote the equivalence class to which g belongs. Now our goal is to turn G/~ into a group. The natural operation that comes to mind is gN * hN = (gh)N -- but we need to ensure that this is well-defined. That is to say, if gN=g'N and hN=h'N (so g~g' and h~h'), then we must have (gh)N=(g'h')N (so gh~g'h'). You can check that the N being normal in G is exactly what we need for this to work. That G/~ is a group is easy to verify (for instance the identity element is eN and the inverse of gN is g-1N), and we call it the quotient group of G by N, which we denote by G/N.

    Yup, that's what I'm suggesting. Just set up the most obvious map from G to G/M x G/N you can think of (remember, the elements of the latter are ordered pairs (gN, gM)), and check that it's an injective homomorphism.

    The abelian group stuff isn't going to be very helpful here at all. (And MN doesn't have to be abelian. For example, S_4 contains C_2 x C_2 and A_4 as normal subgroups, and their product, A_4, is nonabelian.)
  6. Mar 4, 2008 #5
    Okay, so please correct me if I'm wrong about this then:

    Fix any n in N and any m in M and make a homomorphism [itex]\phi:G\rightarrow (G/N\times G/M)[/itex] defined by [itex]\phi(g)=(g n , g m)[/itex]. Can I say for sure that this is injective? Certainly, each component is not (necessarily) injective, that is, f(g)=gn is not injective into gN, because, unless N={e}, |G/N| < |G|, right? Is this where the bit about M and N having only the identity in common comes in?

    Sorry if I seem a bit dense. Your help is very much appreciated, morphism!
  7. Mar 4, 2008 #6
    [itex]\phi(g)=(g n , g m)[/itex] doesn't make sense, elements of [itex]G/N\times G/M[/itex] are ordered pairs of cosets, i think you want [itex]\phi(g)=(gN, gM)[/itex]

    use the definition of injective, and explain where you get stuck, remember g_1N = g_2N iff (g_1)^-1g_2 is in N, likewise for M.
    Last edited: Mar 4, 2008
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