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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?

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# Homework Help: An Abelian Subgroup Problem

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