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

Oct 2009
195
19


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
Jun 2009
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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!
Please take the time to learn some LaTeX so we can help you better.

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. \(\displaystyle mn=nm\).
 
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Oct 2009
195
19
Please take the time to learn some LaTeX so we can help you better.

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. \(\displaystyle mn=nm\).
Ah. Yea, my mistake was pretty stupid.

I see what you did. Thanks very much!