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Here is a proof from Keith Condrad's document with cosets instead of classes:

Theorem 2. If $G$ is a finite group and $N \triangleleft G$ then any element of $G$ with order relatively prime to $[G:N]$ lies in $N$. In particular, if $N$ has index $2$ then all elements of $ G$ with odd order lie in $N$.

Proof: Let $g$ be an element of $G$ with order $m$, which is relatively prime to $[G:N]$. The equation $g^m=e$ gives $(gN)^m=N \in G/N$. Also $(gN)^{[G:N]}=N$, as $[G:N]$ is the order of $G/N$.

So the order of $gN \in G/N$ divides $m$ and $[G:N]$.

These numbers are relatively prime, so $gN=N$, which means $g \in N$.

Why is it required that $G$ is finite? Does the theorem also hold for groups in general? I suspected it because of $[G:N]$, but we set it out to be coprime to $m$, is $[G:N]=\infty$ allowed?

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If the quotient group $G/N$ is finite and $g\in G$ is an element of finite order then it works even if $G$ is infinite. If one of them is infinite then you can't talk about their orders being coprime anyway.

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