# Finding groups from presentations

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?

• Would you consider a group to satisfy these conditions if $a^2 = b^2 = 1$? – Morgan Rodgers Sep 20 '17 at 0:25
• Yes, I would. That would give me the Klein-4 group, right? – Analytical Sep 20 '17 at 0:27

## 1 Answer

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 — $$Q_8$$.