The Fundamental group of Klein Bottle My question is if
$$\pi_{1}(KB)\cong\frac{\mathbb{Z}(a)}{\langle a^{2} \rangle}*\frac{\mathbb{Z}(b)}{\langle b^{2}\rangle}\cong\frac{\mathbb{Z}(a)*\mathbb{Z}(b)}{\langle a^{2}b^{2} \rangle}$$
and
$$\dfrac{\mathbb{Z}(c)}{\langle c^{2} \rangle} \cong \mathbb{Z}_{2},$$
where KB is the Klein Bottle, thus 
$$\frac{\mathbb{Z}(a)}{\langle a^{2}\rangle}*\frac{\mathbb{Z}(b)}{\langle b^{2}\rangle}\cong \mathbb{Z}_{2}*\mathbb{Z}_{2}$$
and
$$\pi_{1}(KB)\cong \mathbb{Z}_{2}*\mathbb{Z}_{2} ?$$
It makes sense to me, but I'm not sure.
Claim: $\pi(K) = \mathbb{Z} \rtimes \mathbb{Z}$

By the application of Van Kampen's Theorem to two dimensional CW complexes we have:

$ \pi(K) = &lt;a,b|abab^{-1}=1&gt; $

Let $A$ be the subgroup generated by $a$ and $B$ be the subgroup generated by $b$.  Then since $bab^{-1}=a^{-1}$, we have that $B$ is a normal subgroup.

We can show that every element has a unique representation in the form $b^na^m$, so that $BA=\pi(K)$ and $B\cap A = 0$.

Now $\pi(K)$ is the internal semidirect product of $A$ and $B$, which are each isomorphic to $\mathbb{Z}$.
We know that $KB$ = $RP^2$ # $RP^2$.So $\pi_1(KB) = \pi_1(RP^2)* \pi_1(RP^2) = Z_2*Z_2.$.