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

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.

closed as off-topic by Eevee Trainer, Najib Idrissi, Yanior Weg, Dietrich Burde, Math1000May 2 at 0:05

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• I just corrected the question. Thank you. – Myown Gait May 1 at 5:24
• Still don't need $aa^{-1}.$ – coffeemath May 1 at 5:30
• aa^{-1} has no effect, right? so That's isomorphic to $\Bbb Z_3$? – Myown Gait May 1 at 5:38
• We don't know what $M$ is! Where is the Heegaard diagram you are talking about? – Najib Idrissi May 1 at 14:36

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 $$\pi_1(M) = \mathbb{Z}/ 3\mathbb{Z}$$.