The Fundamental group of Klein Bottle

Question

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.

Answer 1

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 $\mathbb{Z}$.

Answer 2

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 $\mathbb{Z} \oplus (\mathbb{Z}/2)$).

Answer 3

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 $\mathbb{Z}_{2}*\mathbb{Z}_{2}$.

Answer 4

We know that $KB$ = $RP^2$ # $RP^2$.So $\pi_1(KB) = \pi_1(RP^2)* \pi_1(RP^2) = Z_2*Z_2.$.