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An Abelian Subgroup Problem

  1. Sep 27, 2008 #1
    The problem statement, all variables and given/known data
    Suppose that N and M are two normal subgroups of G and that N and M share only the identity element. Show that for any n in N and m in M, nm = mn.

    The attempt at a solution
    I basically have to show that NM is abelian. Since N and M are normal, it follows that

    [tex]nm = mn_1 = n_1m_1 = m_1n_2 = \cdots,[/tex]

    where [itex]n_1,n_2,\ldots \in N[/itex] and [itex]m_1,m_2,\ldots \in M[/itex]. This is all I know. I don't know how to use the fact that [itex]N \cap M = \{e\}[/itex] or how it plays any role. Any tips?
     
  2. jcsd
  3. Sep 27, 2008 #2

    Dick

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    Sure. Stare at n*m*n^(-1)*m^(-1) for a little bit. If you still don't see it, I'll give you another hint.
     
  4. Sep 27, 2008 #3
    Since M is normal, nmn-1 equals some m' in M and since N is normal, mn-1m equals some n' in N. This yield that m'm-1 = nn'. Aha! Then m'm and nn' must be the identity so m' is the inverse of m and n' is the inverse of n and after some rearrangement, we get that nm = mn.

    Nice. What made you think of looking at nmn1m-1?
     
  5. Sep 27, 2008 #4

    Dick

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    I thought you'd get it written in that form. nm=mn is the same thing as nmn^(-1)m^(-1)=e. It even has a name, that expression is called the commutator. I think of commuting, I think of commutator. Just experience, I guess.
     
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