Let $F$ be freely generated by $\{a, b, c\}$. Prove that $F/N$ is free with basis $aN, bN$, where $N$ is the normal subgroup of $F$ generated by $c$ My question stems from the following exercise in free groups:

Let $F$ be freely generated by $\{a, b, c\}$ and let $N$ be the normal subgroup of $F$ generated by $c$. Prove that $F/N$ is freely generated by $\{aN, bN\}$

The idea of my proof starts by establishing a homomorphism $\theta$ of $F$ into a free group generated by $\{a, b\}$ that sends $c$ to $1$ and keeps $a, b$ intact. If I can show that $N$ = $\operatorname{ker}(\theta)$, the result seems to follow easily in my head.
The first inclusion, $N \subseteq \operatorname{ker}(\theta)$ is trivial by definition. My problem is showing the other way around. The book I'm using says that, if $f \in F$ is not in $N$, then $f = f_1 n_1$, where $n_1 \in N$ and $\mathbf{f_1}$ is a product on $\mathbf{a, b}$.
The part in bold is not clear to me. Why is, for instance, $acb$ in $f_1N$ (for some $f_1$)? The only thing I am able to see about $N$ is that it contains all $x^{-1}c^mx$, where $m \in \mathbb{Z}$ and $x \in F$, and it seems very hard to see that EVERY word containing $c$ can be expressed as $f_1N$... Am I missing something?
Any help is appreciated!
Thanks in advance!

Let $f$ be a reduced word in $a,b,c$ that is not in $N$. You want to show that $f=f_1n$ where $n\in N$ and $f_1$ is a reduced word that does not involve $c$.
Proceed by induction on the sum of the times that $c$ and $c^{-1}$ occur in $f$. If the total is $0$, then we can take $f=f_1$, $n=e$.
Now assume that $f$ has $r\gt 0$ occurrences of $c$ and $c^{-1}$; we can write $f=w_1c^sw_2$, where $w_1$ and $w_2$ are reduced words, $w_2$ does not involve $c$ or $c^{-1}$, $w_1$ does not end in $c$ or $c^{-1}$, and $s\in\mathbb{Z}$, $s\neq 0$. Then we have
$$f = w_1c^sw_2 = w_1w_2c^s[c^s,w_2],$$
where $[x,y]=x^{-1}y^{-1}xy = x^{-1}x^y$.
Since $c^s$ and $[c^s,w_2]=c^{-s}(c^s)^{w_2}$ are both in the normal closure of $\langle c\rangle$, then $c^s[c^s,w_2]\in N$. Thus,
$$fN = w_1w_2c^s[c^2,w_2]N = w_1w_2N.$$
Thus, we can write $f$ as $w_1w_2n_1$, with $n_1\in N$.
But $w_1w_2$ has strictly fewer occurrences of $c$ and $c^{-1}$ than $f$ does, so by induction we can write $w_1w_2=f_1n_2$ for some $f_1$ that does not involve $c$, and $n_2\in N$. Hence $f=w_1w_2n_1 = f_1(n_2n_1)$, with $f_1$ a word that does not involve $c$, and $n_2n_1\in N$, as desired.
