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I am trying to solve the following problem: Find all groups with two generators $a$ and $b$ in which $a^4 = 1, b^2 = a^2, $ and $bab^{-1} = a^{-1}.$

I know that this is a presentation of the quaternion group, and am confused about whether this presentation can be a part of the presentation of some other group. Is it possible that this can be part of the presentation of another group? How would I construct another group?

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    $\begingroup$ Would you consider a group to satisfy these conditions if $a^2 = b^2 = 1$? $\endgroup$ – Morgan Rodgers Sep 20 '17 at 0:25
  • $\begingroup$ Yes, I would. That would give me the Klein-4 group, right? $\endgroup$ – Analytical Sep 20 '17 at 0:27
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Let $Q_8 = \langle\, a, b \mid a^4 = 1,\ b^2 = a^2,\ bab^{-1} = a^{-1} \,\rangle$. For every group $G$ with generators satisfying the relations, there is a surjection from $Q_8$ to $G$ by the von Dyck theorem or by the universal property of group presentation. (See [Aschbacher 2000, (28.6)] or [Grillet 2007, Theorem 7.2].) Thus $G$ is isomorphic to one of

  • trivial group — $1$,
  • group of order $2$$C_2$,
  • Klein four group — $V_4$, or
  • the whole group — $Q_8$.
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