How to show that $\langle a, b \mid a^2, b^2, (ab)^4\rangle$ gives the dihedral group?

Question

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?

Answer 1

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 $D_8$.

Answer 2

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 $\;D_4$.

Answer 3

In general,

$$D_{2n}\cong\langle a,b\mid a^2, b^2, (ab)^n\rangle.\tag{1}$$

I shall prove this here. Call the presentation $P_n$.

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Typically, the dihedral group is defined to be given by

$$D_{2n}\cong\langle r,s\mid r^2, s^n, rsr=s^{-1}\rangle.\tag{2}$$

Let $s=ab$ in $(1)$. Then $b=a^{-1}s=as$. Now, by Tietze transformations, we have

$$\begin{align} P_n&\cong \langle a,b,s\mid a^2, (as)^2, s^n, b=as\rangle\\ &\cong \langle a,s\mid a^2, s^n, asas\rangle\\ &\cong\langle a,s\mid a^2, s^n, asa=s^{-1}\rangle\\ &\stackrel{(2)}{\cong}D_{2n}. \end{align}$$