Let $G$ be a group and let $g \in G$. Let $g $ have infinite order and suppose that $N$ is a finite normal subgroup of $G$. Prove that the element $gN \in G/N$ has infinite order.
So I know that the definition of the order of $g$ is the smallest positive integer such that $g^n = 1$ and if no such $n$ exists then $g$ has infinite order. We're given that $g$ has infinite order so we know that no such positive integer $n$ exists.
I also know that if $N$ is a normal subgroup of $G$ then for all $g\in G, gNg^{-1} = N$.
I think the best way to prove that $gN$ has infinite order would be to prove by contradiction because we are trying to prove that such a positive $n$ doesn't exists. So suppose not.
So my attempt at a proof would be: Assume $gN \in G/N$ has finite order. Then there exists $n$ such that $g^n \in N$. We know $g \in G$ has infinite order. So there does not exist an $n$ such that $g^n = 1$. (I think of this as $g^n \in N/ \{ 1 \}$. $N$ is normal in $G$. So for all $g\in G, gNg^{-1} = N$.
Since $g^n \in N$ and $gNg^{-1} = N$ for all g, then $g^n Ng^{-n} \in N$. But since g is infinite and we said $g^n \in N/ \{ 1 \}$, then $g^nNg^{-n} \neq (1)N(1) = N$? But that's a contradiction?
That's as far as I have. Could I get help with the rest or if there is another way to prove this, that would be very helpful.
Best Answer
An idea: Suppose $\;|N|= n<\infty\;$ , and suppose $\;|gN|=k\in\Bbb N\;$ , then
$$N=(gN)^k=g^kN\iff g^k\in N\implies g^{kn}=1$$
and now get your contradiction.