Prove that if $G = \langle a, b \mid a^4, a^2b^{-2}, aba^{-1}b \rangle$, then $G \cong Q_8$.

I started by trying to define a homomorphism $\varphi: F(a,b) \to Q_8$ by $\varphi(a) = i$, $\varphi(b) = j$ and $\varphi(ab) = k$.

EDIT: this is not true (Since $Q_8$ is abelian, I know that the commutator $aba^{-1}b^{-1} \in Ker(\varphi)$),

So the elements $a^4, b^4, (ab)^4$ are in the kernel. However, I can't seem to show that this kernel is equal to the normal closure of the group $\langle a^4, a^2b^{-2}, aba^{-1}b \rangle$. Is there an easier way to do this problem? We just proved van Dyck's theorem in class, and our professor hinted that it applies to this problem, but I can't see how to use it.

Any hints will be appreciated, but I'd rather not have a complete solution.

  • $\begingroup$ But $Q_8$ is not abelian. $\endgroup$ – Derek Holt Nov 12 '14 at 1:14
  • $\begingroup$ @DerekHolt Almost... I thought I went over the whole multiplication table... I edited the question to account for that. The rest of it still stands though. $\endgroup$ – Johanna Nov 12 '14 at 1:20

The second relation gives $b^2 = a^2$. This says $|a| = |b|=4$, and that the elements $a^n b^m$, with $m\ge 3$ can be reduced to $a^{r}b^{s}$ but with $r=\{0,1,2,3\}, s = \{0,1\}$

The third relation: $aba^{-1}b = 1 \implies ab = b^{-1}a \implies bab =a$.

Moreover $bab = a \implies ba = ab^{-1} = ab^3 = a(b^2)b = a(a^2)b =a^3b$, ie, $ba=a^3b$. So elements with mixed letters like $ba, bab, etc$ can also be expressed as $a^rb^s$ with $r=\{0,1,2,3\}, s= \{0,1\}$

This means, $G$ has 8 distinct elements: $$1, a, a^2, a^3$$ $$b, ab, a^2b, a^3b$$

Since $i,j \in Q_8$ satisfies the relations in $G$, Von Dyck's Theorem gives a surjective homomorphism $G \to Q_8$, $a \mapsto i, b\mapsto j$

We have the following:

  • $|G| = 8$
  • The homomorphism is also an injection (same group order + surjectivity)
  • The homomorphism is now a bijection, hence an isomorphism.

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