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

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

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!

2. Originally Posted by 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!

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. $\displaystyle mn=nm$.

3. Originally Posted by Bruno J.
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. $\displaystyle mn=nm$.