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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.

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  • $\begingroup$ What are you denoting by $\Bbb Z(a)$? I suppose you mean the infinite cyclic group. You could write $\langle a \rangle$ and say $a$ has infinite order, or just write $\Bbb Z/2\Bbb Z$. $\endgroup$ – Pedro Tamaroff Mar 28 '15 at 20:59
  • $\begingroup$ You are correct, it's the infinite cyclic group. $\endgroup$ – altairsot Mar 28 '15 at 21:03
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    $\begingroup$ The arithmetic that you do with quotients and free products in the first display is not valid. $\endgroup$ – Lee Mosher Mar 28 '15 at 21:04
  • $\begingroup$ And anyway, $\mathbb{Z}_2 * \mathbb{Z}_2$ contains $\mathbb{Z}$ with finite index whereas $\pi_1(KB)$ contains $\mathbb{Z} \oplus \mathbb{Z}$ with finite index, and these are incompatible. $\endgroup$ – Lee Mosher Mar 28 '15 at 21:05
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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) = <a,b|abab^{-1}=1> $

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 $\mathbb{Z}$.

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The group $$ G_1 = \langle a, b |a^2b^2=1 \rangle $$ is not isomorphic to the group $$ G_2 = \langle c, d | c^2=1 , d^2=1 \rangle. $$ 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 $\mathbb{Z} \oplus \mathbb{Z}/2$).

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

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

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    $\begingroup$ Van Kampen's theorem does not imply that $\pi_1(A \, \# \, B) = \pi_1(A) * \pi_1(B)$. $\endgroup$ – Lee Mosher Mar 29 '15 at 15:23

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