I am trying to find the Cayley Graph the group $$G=\langle a,b\mid a^2,b^2,(ab)^2\rangle.$$
It is easy to prove that $G$ is isomorphic to the dihedral group $D_4$, and $G$ must have $8$ elements.
I am not able to find all of the elements.
Any help would be appreciated
Lemma: For any $x$ in a group $H$, we have $$x^2=e\iff x=x^{-1}.$$
Proof: Let $x\in H$. Suppose $x^2=e$. Multiply by $x^{-1}$ on, say, the left. Then
$$\begin{align} x^{-1}&=x^{-1}e\\ &=x^{-1}x^2\\ &=(x^{-1}x)x\\ &=ex\\ &=x. \end{align}$$
Conversely, suppose $x=x^{-1}$. Multiply on, say, the right by $x$. Then $x^2=xx=x^{-1}x=e$. $\square$
Since $a^2=b^2=(ab)^2=e,$ we have
$$\begin{align} ab&=(ab)^{-1}\\ &=b^{-1}a^{-1}\\ &=ba, \end{align}$$
so that, via Tietze transformations, the presentation is isomorphic to
$$\langle a,b\mid a^2, b^2, ab=ba\rangle,$$
which is a presentation for the Klein four group (as it is the direct product of $\Bbb Z_2$ with itself), the Cayley graph for which is easy to find.
It should look like this:
This image was found here.
The red arrows are multiplying by $a$; the green arrows, by $b$; and the blue, $ab$.
