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.
 A: The group
$$
G_1 = \langle a, b \mid a^2b^2=1 \rangle
$$
is not isomorphic to the group
$$
G_2 = \langle c, d \mid c^2=1 , d^2=1 \rangle.
$$
Proof: they have different abelianizations. Indeed, it is clear that the abelianization of $G_2$ is $(\mathbb{Z}/2)^2$. On the other hand, $G_1 \twoheadrightarrow \mathbb{Z}$ by the rule $a \mapsto 1$, $b \mapsto 0$, so the abelianization of $G_1$ can't be finite (in fact it is $\mathbb{Z} \oplus (\mathbb{Z}/2)$).
A: The fundamental group of the Klein bottle is torsion free. So it cannot contain any copies of $\mathbb{Z}_{2}$ . So it cannot be isomorphic to 
$\mathbb{Z}_{2}*\mathbb{Z}_{2}$.
A: Claim: $\pi(K) = \mathbb{Z} \rtimes \mathbb{Z}$
By the application of Van Kampen's Theorem to two dimensional CW complexes we have:
$$\pi(K) = \langle a,b \mid a b a b^{-1}=1 \rangle.$$
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 $\mathbb{Z}$.
A: We know that $KB$ = $RP^2$ # $RP^2$.So $\pi_1(KB) = \pi_1(RP^2)* \pi_1(RP^2) = Z_2*Z_2.$.
