Fundamental group of 2-fold projective plane and Klein bottle

Question

By the classification of surfaces, I know that the connected sum of two projective planes is the Klein bottle. So its fundamental group should be $\langle a,b \mid a^{2}b^{2} = 1\rangle$ (according to classification of surfaces section in Munkres' Topology). But I computed the fundamental group of the Klein bottle using the van Kampen theorem and it is coming out to be $\langle a,b \mid aba^{-1}b = 1\rangle$. I am not able to see where am I going wrong. Any help would be appreciated.

Answer 1

I am basically expanding on Thomas's answer for future viewers. What is being used is known as "Tietze Transformations" which is a standard way in combinatorial group theory to go from one group presentation to another group presentation such that in each step the underlying group remains the same. Here's one instance where Tietze Transformation is being used : Tietze Transformations: $\langle a, b, c\mid b^2, (bc)^2\rangle$ and $\langle x, y, z\mid y^2, z^2\rangle$.

We are going to apply Tietze Transformations on the following group: $$G = \langle a,b \mid a^{2}b^{2} = 1 \rangle$$ Apply (T3) to add a generator $$G = \langle a,b,x \mid a^{2}b^{2} = 1, x = ba \rangle$$ Apply (T1) to add the relation $a^{-1}xa = x^{-1}$ derived from the relations already present $$G = \langle a,b,x \mid a^{2}b^{2} = 1, x = ba, a^{-1}xax = 1 \rangle$$ Apply (T3) to add the generator $y$ $$G = \langle a,b,x,y \mid a^{2}b^{2} = 1, x = ba, a^{-1}xax = 1, y = a^{-1} \rangle$$ Apply (T1) to add the relation $yxy^{-1}x = 1$ $$G = \langle a,b,x,y \mid a^{2}b^{2} = 1, x = ba, a^{-1}xax = 1, y = a^{-1}, yxy^{-1}x = 1 \rangle$$ Apply (T2) to delete the relation $a^{-1}xax = 1$ $$G = \langle a,b,x,y \mid a^{2}b^{2} = 1, x = ba, y = a^{-1}, yxy^{-1}x = 1 \rangle$$ Apply (T4) to remove the generator $a$ since $a = y^{-1}$ $$G = \langle b,x,y \mid y^{-2}b^{2} = 1, x = by^{-1}, yxy^{-1}x = 1 \rangle$$ Apply (T4) again to remove $b$ since $b = xy$ $$G = \langle x,y \mid y^{-2}(xy)^{2} = 1, yxy^{-1}x = 1 \rangle$$ Observe that the relation $y^{-2}(xy)^{2} = 1$ can be derived from $yxy^{-1}x = 1$. Hence applying $(T2)$ to remove the relation $y^{-2}(xy)^{2} = 1$ we have $$G = \langle x,y \mid yxy^{-1}x = 1 \rangle$$

Answer 2

These two groups are isomorphic.

If $G=\langle a,b, a^2=b^{-2}\rangle$, let $x= ba$, then $a^{-1} x a= a^{-1} b a^2= a^{-1}b^{-1}= x^{-1}$. Let $A=a^{-1}, B=x$, we have $ABA^{—1}=B^{-1}$. It is not difficult to see that this is a new presentation of the same group.