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I am trying to find a better solution to the following problem.

Show that the group $Q_8 = \langle a, b : a^4 = 1, a^2 = b^2, bab^{-1} = a^{-1}\rangle$ has order $8$.

My attempt: the last relation gives $ba = a^3 b$, so every element can be written on the form $a^i b^j$. The other relations restrict it further: $a^i b^j$ with $i \in \{ 0,1 ,2 ,3 \}$ and $j \in \{0, 1\}$. This gives at most eight elements, so it suffices to show they are different. The powers of $a$ are distinct since $a$ has order $4$, and the elements $a^kb$ are distinct for the same reason.

My question is: how can I efficiently show that $a^kb \neq a^{\ell}$? The case $k = \ell$ is easy: we get $b = 1$, which contradicts the second relation. If $k > \ell$, we get $ab = 1$ or $a^2 b = 1$ etc. I think I went through all the possible cases, but it felt really ad-hoc. Is there a better way to prove this?

Shaun
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hdecristo
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    If $a^k b = a^l$, then $b=a^{-k}a^l\in \langle a\rangle$ and $Q_8$ would be cyclic. – Anonymous IV Jun 21 '25 at 14:44
  • Note that just because there is a relation $a^4=1$, does not imply that the order of $a$ is $4$. – Hagen von Eitzen Jun 21 '25 at 15:39
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    To expand in my comment: The slightly different group $\langle,a,b: a^n=1,a^3=b^3, bab^{-1}=a^{-1},\rangle$ is cyclic of order $3$ if $n$ is odd (with $a$ having order $1$) and is cyclic of order $6$ if $n$ is even (with $a$ having order $2$), In general, the order of $a$ is not $n$. – Hagen von Eitzen Jun 21 '25 at 15:52
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    @AnonymousIV Yes, but why is that a contradiction? How do you know that it is not cyclic? – Derek Holt Jun 22 '25 at 21:37
  • @DerekHolt That's a good point. Suppose the group is cyclic. Then the third relation gives $|a|=1$ or $2$. I think you can't suppose $a=1$ because that does not follow from the given relations. Similarly, you can't have $b$ be a generator of order $4$ ($a = b^2$ and second relation would give $b^2 = 1$). However, I don't see why the group could not be, for example, the Klein four group. – hdecristo Jun 22 '25 at 22:23

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The complex matrices $A = \left(\begin{array}{rr}0&1\\-1&0\end{array}\right)$ and $B = \left(\begin{array}{rr}i&0\\0&-i\end{array}\right)$ satisfy the group relations $A^4=I$, $A^2=B^2$, $BAB=A^{-1}$, so defining $H$ to be the multiplicative group generated by $A$ and $B$, there is an epimorphism $G \to H$ mapping $a \mapsto A$ and $b \mapsto B$, and hence $|G| \ge |H|$..

Since $A$ has order $4$ and $A$ and $B$ do not commute, the group $H$ must have order at least $8$, so $|G| \ge 8$. But you have already shown that $|G| \le 8$, so $|G|=8$, and the eight products $a^ib^j$ all represent distinct elements of $G$.

Derek Holt
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  • Thank you! But sorry, I'm still a bit confused. Why is there such an epimorphism? If you had picked $A = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix}$ and $B = \begin{pmatrix} -1 & 0 \ 0 & 1 \end{pmatrix}$, would there be an epimorphism from the Klein four group onto $G$? – hdecristo Jun 24 '25 at 01:54
  • The existence of the epimorphism $G \to \langle A,B \rangle$ follows directly from a basic theorem on group presentations that you really need to be familiar with to have any hope of tackling problems like this. As for your second question, this result says that there is an epimorphism from $G$ to the Klein four-group (not the other way round). – Derek Holt Jun 24 '25 at 19:54
  • Oh, I see. Thank you! – hdecristo Jun 24 '25 at 20:04