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Let $F_{3,0}$ denote the $2$-punctured disk (or $3$-punctured sphere). I've seen in many textbooks that $\pi_1(F_{3,0}\times S^1)=\langle a_1,a_2,a_3,h| a_ih=ha_i,\ 1\leq i\leq 3,\ a_1a_2a_3=1 \rangle$ but I do not know how to prove it and I do not know from where to start.

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    $\begingroup$ I'm not sure about this presentation, but the fundamental group is just $F(a,b) \times \mathbb{Z}$ where $F(a,b)$ is the free group on two generators. $\endgroup$ – Connor Malin Dec 10 '19 at 3:01
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First of all $\pi_1(X\times Y)\simeq\pi_1(X)\times\pi_1(Y)$ as you can see here.

Now $\pi_1(S^1)\simeq\mathbb{Z}$ as we all know. On the other hand your $F_{3,0}$ is homotopy equivalent to the wedge sum of two circles $S^1\vee S^1$ and so $\pi_1(F_{3,0})\simeq\mathbb{F}(a_1,a_2)$ where on the right side we have the free group on two elements. And so

$$\pi_1(F_{3,0}\times S^1)\simeq \mathbb{F}(a_1,a_2)\times\mathbb{Z}$$

The group on the right side has the following presentation (which you can derive from here):

$$\langle a_1,a_2,h\ |\ a_ih=ha_i, i=1,2 \rangle$$

This is the same presentation that you have, simply because your last relation means $a_3=(a_1a_2)^{-1}$ and so $a_3$ is fully expressible via other generators. Meaning both the generator and the relation can be removed from the presentation.

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