How to show that this group presentation is isomorphic to dihedral group? The group with the presentation $\langle a, b \mid a^2, b^2, (ab)^4\rangle$ seems like a dihedral group of order 8. But I fail to explain why. Could anyone show me how to prove it?
 A: I will assume the standard presentation of the dihedral group of order 8 as generated by a rotation $r$ and a reflection $s$, namely, $D_8 = \langle r, s | r^4=s^2=1, rs = sr^{-1}\rangle$.
Let $G = \langle a, b | a^2, b^2, (ab)^4 \rangle$ . We can define a homomorphism $\varphi : G \rightarrow D_8$ by $\varphi(a) = s$ and $\varphi(b) = sr$. This really does define a homomorphism because $$(\varphi(a))^2 = (\varphi(b))^2 = (\varphi(a)\varphi(b))^4 = 1,$$ so all relations in $G$ are satisfied. Likewise, we may define the homomorphism $\psi : D_8 \rightarrow G$ by $\psi(r) = ab$ and $\psi(s) = a$. All relations in $D_8$ are satisfied. For instance, $$\psi(r)\psi(s) = aba = ab^{-1}a^{-1} = a(ab)^{-1} = \phi(s)(\phi(r))^{-1}.$$
Now, $\varphi$ and $\psi$ are inverses of each other. (We just check for the generators: $\psi \circ \varphi(a) = a$, etc.) Thus they are isomorphisms, and so $G$ is isomorphic to $D_8$.
A: Yes, it is the dihedral of order eight. Just observe that
$$s=a\;,\;\;t=ab\implies sts=a(ab)a=ba=(ab)^{-1}=t^{-1} $$
and you get the usual presentation for $\;D_4$.
