# Why does $G$ need to be finite in this proof of a normal subgroup $N$ containing all elements of order coprime to $[G:N]$?

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?

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.