Presentation of Klein group: Show $G=\langle a,b\mid a^2,b^2,(ab)^2,(ba)^2,ab^2a\rangle$ is iso to $K_4=\langle x,y\mid x^2,y^2,xyx^{-1}y^{-1}\rangle$

Question

Let

$$G = \langle a, b \mid a^2, b^2, (ab)^2, (ba)^2, a b^2 a \rangle.$$

I need to show that this group is isomorphic to Klein Group

$$K_4 = \langle x,y\mid x^2, y^2, xyx^{-1} y^{-1}\rangle.$$

I prove that this group has order $4$, but I don't know if $G$ is cyclic or not. I would like to know if anyone could give me any hint to show this using Tietze transformations. I try substitute $ab$ by $x$ and remove one relation, like this $$ x = ab \Rightarrow b = a^{-1}x \Rightarrow (ab)^2 = x^2 \Rightarrow (ba)^2 = (a^{-1} x a)^2 = a^{-1} x^2 a \backsim x^2 $$ So $G$ is isomorphic to $$ \langle a, x \mid a^2, x^2, (a^{-1}x)^2,x a^{-1} x a\rangle $$ From here, I tried everything and I couldn't get to the representation of Klein Group.

Answer 1

Since $(ab)^2=abab=aba^{-1}b^{-1}$ (as $a^{-1}=a$ and $b^{-1}=b$), we have that the group given by $G$ is isomorphic to a quotient of $K_4$. But $(ba)^2$ is equivalent to $(ab)^2$ (why?) and, since $b^2=1$, we have that $ab^2a=a^2=1$ follows from existing relations, so each can be eliminated by Tietze transformations. Hence the groups are isomorphic.