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How to show that if $N \ \& \ M$ are 2 normal subgroups of group $G$ and $N\cap M=\{e\}$ (identity element), then for any $n\in N \ \&\ m\in M $, $nm=mn$?

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    $\begingroup$ Please, try to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$ May 14, 2016 at 6:31
  • $\begingroup$ Similar question: math.stackexchange.com/questions/253131/… (Although probably not a duplicate, since the other question asks about explanation of one specific step in the proof provided by the OP.) $\endgroup$ May 14, 2016 at 6:34

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If we can show that $m^{-1}nmn^{-1}=e$, then multiplying by $m$ and $n$ from the respective sides you'll get the desired result. Now, regarding the element $m^{-1}nmn^{-1}$: use the normality of $M$ and $N$ to show that it lies in both subgroups.

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