Proving infinite order. 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. 
 A: 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.
A: The end of your proof does not really work (as your question marks there indicate you already suspect). You are too much obsessed with what normality means, but all you need it for in this exercise is just to know that the quotient $G/N$ is actually a group. Also, notice that you did not use the fact that $N$ is finite, so you should be suspect of your proof. 
So, you know that $g^n\in N$, you also know that $g^m\ne 1$, for all $m>0$, and you know that $N$ is finite. Now write $h=g^n$ and consider all powers $h^k$, $k\ge 0$. Can any two such powers be equal? Do they all belong to $N$? Does that agree with $N$ being finite?
Your proof is not necessarily the shortest one, but it is certainly very close to being correct. My answer follows your thoughts. You may wish to find a shorter solution if you like.
A: An alternative proof: Suppose $|gN|=m$. Then the set $\{g^kN\}_{k=0}^{\infty}$ of cosets of $N$ associated with powers of $g$ must have exactly $m$ elements. Namely, the set is equal to $\{N,gN,g^2N,\ldots,g^{m-1}N\}$. But the set
$$\bigcup_{k=0}^{m-1}{g^kN}$$
has only finitely many elements (to be specific, it has $m|N|$ elements), so it cannot contain all powers of $g$.
A: Let $H=\langle g\rangle$; then $HN$ is an infinite subgroup of $G$ containing $N$ as a normal subgroup, so it's not restrictive to assume $G=HN$. By the homomorphism theorem,
$$
G/N=HN/N\cong H/(H\cap N)
$$
Since $H$ is generated by $g$, also $H/(H\cap N)$ is cyclic. But $H$ is infinite cyclic and the only finite subgroup of $H$ is $\{1\}$. Thus $G/N$ is infinite as well and, since it is generated by $gN$, this element has infinite order.
