Proving Commutativity in Normal Subgroups with Abelian Subgroup Problem

[e(ho0n3]

Homework Statement
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

$$nm = mn_1 = n_1m_1 = m_1n_2 = \cdots,$$

where $n_1,n_2,\ldots \in N$ and $m_1,m_2,\ldots \in M$. This is all I know. I don't know how to use the fact that $N \cap M = \{e\}$ or how it plays any role. Any tips?

 

[Dick]

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.

 

[e(ho0n3]

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?

 

[Dick]

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.