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This has eluded me, and my professor didn't really explain it well. I understand that given that N is a normal subgroup, that G/N = gN such that g is in G. However I don't really see how this can make N the identity element of G/N

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The identity element in $G/N$ is the class $eN$, which is simply $N$.

More generally, you can show that $$hN=eN \Leftrightarrow h \in N$$

This means that all elements in $N$, and only elements of $N$, become the identity in $G/N$.

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    Oh, so it's because when multiplying any element in N by the entire subgroup, N is the result, so therefore any nN is always just going to be N, and therefore any element of n is always going to return N when you multiply it together by the entire subgroup?2017-02-28
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    @user3491700 Exactly. Moreover, if $hN=N$, then since $e \in N$ you get $he \in hN=N$. This also gives you the converse of your statement.2017-02-28