On the quotient group $\pi_{1}(K)/N$ for the Klein bottle $K$

On the quotient group $\pi_{1}(K)/N$ for the Klein bottle $K$

I know that the Klein bottle $K$ is obtained from the unit square by
making identifications on the boundary with the appropriate directional
arrows. Usually, what is done is that we identify the point $(0,t) \sim
(1,1-t)$ and the point $(t,0) \sim (t,1)$ $\forall t \in [0,1]$.
Now suppose we let $a \in \pi_{1}(K)$ (fundamental group of $K$) be
represented by the path $\gamma(t)= [(0,t)]$, where $[(x,y)]$ is the
element of $K$ represented by $(x,y) \in I^2$. Similarly, we let $b \in
\pi_{1}(K)$ be represented by the path $\delta(t)= [(t,0)]$. Now I know
that $\pi_{1}(K)$ has the presentation $\langle a, b $ $\vert$ $bab^{-1}a=
1 \rangle$. If we let $N= \langle a \rangle= \{a^k: k \in \mathbb{Z}\}$,
since $bab^{-1}= a^{-1} \in N$, it follows that $N$ is a normal subgroup
of $\pi_{1}(K)$. Does this seem about right so far?
My question is that what then is the quotient group $\pi_{1}(K)/N$? I'm
not sure if it is infinite cyclic but I think it has got to be generated
by $b$. I would like to know how I can prove this claim. I encountered
this question on an old topology prelim from about five years ago. Looks
like I need to brush up on some group theory!