# [SOLVED] N,M normal, (N n M) = {e}, then nm = mn

#### davismj

Is this correct? I'm pretty sure it is, but the question was "starred," so I want to make sure its not much harder than I think it is.

Thanks!

#### Bruno J.

MHF Hall of Honor
Is this correct? I'm pretty sure it is, but the question was "starred," so I want to make sure its not much harder than I think it is.

Thanks!

It's not correct; how do you justify the step $$\displaystyle (nN\cap nM)m=m(nN\cap nM)$$?

Let $$\displaystyle C=m^{-1}n^{-1}mn$$. Then $$\displaystyle C = (m^{-1}n^{-1}m)n \in N$$, and $$\displaystyle C=m^{-1}(n^{-1}mn) \in M$$. Therefore $$\displaystyle C=1$$, i.e. $$\displaystyle mn=nm$$.

Aki and davismj

#### davismj

It's not correct; how do you justify the step $$\displaystyle (nN\cap nM)m=m(nN\cap nM)$$?
Let $$\displaystyle C=m^{-1}n^{-1}mn$$. Then $$\displaystyle C = (m^{-1}n^{-1}m)n \in N$$, and $$\displaystyle C=m^{-1}(n^{-1}mn) \in M$$. Therefore $$\displaystyle C=1$$, i.e. $$\displaystyle mn=nm$$.