Solution
= Solution
Every $b^m$ lies in $N$ because $(bN)^m=N$. Conversely choose $r$ with $mr\equiv1\pmod n$; for $a\in N$, $a=(a^r)^m$ because $a^n=e$.
= Solution
Every $b^m$ lies in $N$ because $(bN)^m=N$. Conversely choose $r$ with $mr\equiv1\pmod n$; for $a\in N$, $a=(a^r)^m$ because $a^n=e$.