If $f\colon G\rightarrow H$ is a surjective homomorphism of groups, and $N$ is a normal subgroup of $G$, prove that $f(N)$ is a normal subgroup of $H$.
I know that to prove normal subgroup I must show that it first is a subgroup (proving closure and inverses) and then show that it is normal.
So far I have the following: Subgroup (1) Closure Let $a,b\in f(N)$ $f(a) = e_h$ (identity in $H$) and $f(b) = e_h$ $f(ab)=f(a)f(b)=e_h\cdot e_h=e_h$ therefore, $ab\in f(N)$ (2) Inverses Let $n\in f(N)$ then $f(n) = e_h$ since $n\in N$ and $N\subseteq H$ we can say $n^{-1}\in H$ and $f(n^{-1})=f(n)^{-1}=e_h^{-1}=e_h$ so $f(n^{-1}) = e_h$, therefore $n^{-1}\in N$.
Therefore $N$ is subgroup of $H$.
To show normality of the subgroup: let $a\in H$ and $n\in N$ $f(a^{-1}na)=f(a^{-1})f(n)f(a)=f(a^{-1})e_hf(a)=f(a^{-1})f(a)=e_h$ so $a^{-1}na\in N$.
Therefore $N$ is normal.
We can concluded that $f(N)$ is a normal subgroup of $H$.
Question: Is this on the right track or am I way off for proving that $f(N)$ is a normal subgroup of $H$?
Your proof isn't right because you're assuming that $f(a)=f(b)=e_h$, which isn't necessarily true.
First of all, $f(N)=\{f(x):x\in N\}$. Now, let's pick $a,b\in f(N)$, then $a=f(x)$ and $b=f(y)$, for some $x,y\in N$. Since $f$ is a group homomorphism, $ab=f(x)f(y)=f(x\cdot y)\in f(N)$, because $x\cdot y\in N$. On the other hand, $af(x^{-1})=f(x)f(x^{-1})=f(x\cdot x^{-1})=f(e_{G})=e_{H}$, thus $a^{-1}=f(x^{-1})$ because $a^{-1}$ is unique, then $a^{-1}\in f(N)$ since $x^{-1}\in N$. This proves that $f(N)$ is a subgroup of $H$.
Now, let's take $z\in H$. We want to show that $zf(N)z^{-1}\subseteq f(N)$. In order to prove that we have to show that $zaz^{-1}\in f(N)$ for every $a\in N$. Since $a\in N$, $a=f(x)$, for some $x\in N$. Besides $f$ is surjective means that there is $c\in G$ such that $f(c)=z$, then $$zaz^{-1}=f(c)f(x)f(c^{-1})=f(c\cdot x\cdot c^{-1})\in f(N)$$ because $c\cdot x\cdot c^{-1}\in N$ since $N$ is a normal subgroup of $G$. Hence, $f(N)$ is a normal subgroup of $H$.