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Is it true that $\pi_1(M)=\langle a\mid aaa,aa^{-1} \rangle \cong \Bbb Z_3$?

This is the fundamental group of a three dimensional manifold determined by

its Heegaard diagram.

I'd appreciate your help.

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closed as off-topic by Eevee Trainer, Najib Idrissi, Yanior Weg, Dietrich Burde, Math1000 May 2 at 0:05

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  • $\begingroup$ I just corrected the question. Thank you. $\endgroup$ – Myown Gait May 1 at 5:24
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    $\begingroup$ Still don't need $aa^{-1}.$ $\endgroup$ – coffeemath May 1 at 5:30
  • $\begingroup$ aa^{-1} has no effect, right? so That's isomorphic to $\Bbb Z_3$? $\endgroup$ – Myown Gait May 1 at 5:38
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    $\begingroup$ We don't know what $M$ is! Where is the Heegaard diagram you are talking about? $\endgroup$ – Najib Idrissi May 1 at 14:36
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Unless I know what $M$ is, I can't tell you if this is the correct fundamental group, but I can verify that $$ \pi_1(M) = \langle a | aaa \rangle \cong \mathbb{Z}_3 $$ As others have pointed out, $aa^{-1}$ does not need to be specified as equal to the identity (denoted by $e$ for this answer), since this is implicit in the definition of a group.

Consider the homomorphism $\phi : \mathbb{Z} \to \pi_1(M)$ defined by $n \mapsto a^n$. Suppose $\phi(n) = e$. Then $a^n = e$. Using the division algorithm, we find $n = 3q + r$ for some integers $q$ and $r$ with $0 \leq r < 3$. Thus, $$e = a^n = a^{3q + r} = a^{3q}a^r = (a^3)^q a^r = e^q a^r = e a^r = a^r$$ Therefore, either $r = 0$ or $a = e$ or $a^2 = e$. By the definition of $\langle a | aaa \rangle$, $a \neq e$ and $a^2 \neq e$. Thus, $r = 0$. This means $n = 3q$ for some integer $q$. We have shown $\phi(n) = e$ implies $n = 3q$. Clearly $$\phi(3q) = a^{3q} = (a^3)^q = e^q = e$$ Therefore, $\phi(n) = e$ if and only if $n = 3q$. Thus, $\ker \phi = 3\mathbb{Z}$. By the isomorphism theorem, $\text{im }\phi \cong \mathbb{Z} / \ker \phi$. Clearly $\text{im } \phi = \pi_1(M)$, so we have $\pi_1(M) = \mathbb{Z}/ 3\mathbb{Z}$.

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