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?
[Math] How to show that this group presentation is isomorphic to dihedral group
group-theory
Related Solutions
If $G$ is finite and has generators $x,y$ of order 2, then the elements of $G$ are $x,xy,xyx,xyxy,xyxyx,\dots$ and $y,yx,yxy,yxyx,yxyxy,\dots$ and as soon as you know the first term in those lists to give you the identity element, you're done. It can't be an element like $xyxyx$, because if that's the identity then you multiply left and right by $x$ to find $yxy$ is the identity, and you multiply left and right by $y$ to find $x$ is the identity. So the defining relation must be $(xy)^m=1$ for some positive integer $m$ (note that $(yx)^m=1$ if and only if $(xy)^m=1$).
So your presentation is $$\langle x,y\mid x^2,y^2,(xy)^m\rangle$$ and you seem happy to accept that as dihedral.
Recall that when we say that $G = \langle a, b \, | \, a^2 = b^2 = (ab)^n = 1\rangle$, what we mean is that $G$ is the quotient of the free group $\langle a, b\rangle$ by the normal subgroup $N$ generated by $a^2, b^2, (ab)^n$. Now, let's concretely view $D_n$ as the group of rotations and reflections of the regular $n$-gon which preserve the vertices. I'll assume you're familiar with this group.
We can define a group homomorphism $\varphi :\langle a,b\rangle \to D_n$ by sending $a$ and $b$ to "adjacent" reflections. By this, I simply mean that $\varphi(ab) = \varphi(a)\varphi(b)$ should be a rotation of order $n$. Using our knowledge of $D_n$, it's easy to confirm that $a^2, b^2$ and $(ab)^n$ are in the kernel of $\varphi$. Therefore all of $N$ is contained in the kernel. It follows that there is an induced group homomorphism $$\overline{\varphi}: \langle a, b \, | \, a^2 = b^2 = (ab)^n = 1\rangle \to D_n$$ by the univeral property of the quotient. Moreover, you have shown that the domain of the map has at most $2n$ elements, and by construction $\overline{\varphi}$ is surjective (since $\varphi$ is). Since $|D_n| = 2n$ as well, $\overline{\varphi}$ must be bijective, so we're done.
Best Answer
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 $D_8$.