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Suppose there exists an element in $G/N$ of order $p$, where $p$ is a prime number. In other words, there exists an $a\in G$ such that $[a]^p=[e]$, where $[a]\neq[e]$.

Then why $a^p$ belongs to $N$?

I understand if $a^p$ belongs to $N$, then since $[a^p]=Na^p$, by closure of $N$, $[a^p]$ is a subset of $N$; Since $N=Na^{-p}a^p$, by the fact that $N$ is closed under inversion and multiplication, $N$ is a subset of $[a^p]$, thus $[a^p]=N=[e]$. But although the statement that $a^p$ belongs to $N$ leads to the fact that $[a]^p=[e]$, it's not a necessary condition for these two to be equal.

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oops, sorry, i'll try my best –  pxc3110 Apr 3 at 0:59

2 Answers 2

up vote 4 down vote accepted

You're overcomplicating things.

By definition, $[a]^p=[a^p]$ and saying $[a]^p=[e]$ becomes $$ [a^p]=[e] $$ or $a^p\in N$, because the equivalence relation you're using is $$ x \sim y \text{ if and only if } xy^{-1}\in N. $$ In different notation, where we use $[x]=Nx$, $$ Nx=Ny\text{ if and only if }xy^{-1}\in N. $$

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$[a^p] = [a]^p = [e] \iff a^pN = eN = N$. We also have $a^p = a^p\cdot1 \in a^pN = N$, so $a^p \in N$.

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