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I have the group presentation

$$ G := \langle a,b |a^8=b^8=1,a^{-1}ba=b^{-1},b^{-1}ab=a^{-1} \rangle. $$

Which group is it? Notably, what about its order?

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  • $\begingroup$ All of the elements can be expressed as $a^nb^n$ with $n\in \{0,1,2,3,4,5,6,7\}$. What about products like $ba?\cdots b = ab^{-1}a^{-1}$ and $ba = ab^{-1} = ab^7$ $\endgroup$ Commented Feb 15, 2018 at 23:17
  • $\begingroup$ Yes, all of the elements are of the form $a^n b^n$. Moreover (and if do not make mistakes) $a^2 = b^{-2}$. Does it mean that $a^2$ is central?I'm getting confused all the day about this problem. With my computations |G|=16, but it seems to me this group is not in the table of small groups. I made some mistake probably, but I don't find it. $\endgroup$ Commented Feb 15, 2018 at 23:25
  • $\begingroup$ Find "the" group? Generally the trivial group satisfies a presentation of this kind, with $a=b=1$. $\endgroup$ Commented Feb 15, 2018 at 23:25
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    $\begingroup$ Using $b^{-1}=a^{-1}ba$ in the fourth relation I get $(a^{-1}ba)ab=a^{-1}$, $baab=1$, and then $a^2=b^{-2}$. $\endgroup$ Commented Feb 15, 2018 at 23:34
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    $\begingroup$ @MarkBennet That is not how presentations work. Formally, each presentation does define a unique group. (See en.wikipedia.org/wiki/Presentation_of_a_group) $\endgroup$ Commented Feb 16, 2018 at 0:13

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This is the quaternion group, of order $8$. First, note that $a^2=b^{-2}$, as you've noticed. In particular, $a^2$ is centralised by $b$, but $b$ also inverts $a$ by conjugation, so we find $a^2=(a^2)^b=a^{-2}$, and so $a^4=1$. Now, $\langle a\rangle$ is normal in $G$, has order at most $4$, and the quotient has order at most $2$ (since $b^2\in\langle a\rangle$). This shows the group has order at most $8$, but $Q_8$ clearly satisfies the presentation (with $a=i$ and $b=j$, for example).

Note that we never use the first two relations! Indeed, $\langle a,b\mid a^b=a^{-1}, b^a=b^{-1}\rangle$ is already the quaternion group! (This is often useful to remember.)

(I'm using $a^b$ throughout to denote the conjugate of $a$ under $b$, that is $a^b=b^{-1}ab$.)

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Consider the permutation group $Q = \langle \alpha = (1234)(5687), \beta = (1538)(2746)\rangle$. It's easy to check that the mapping $a\mapsto\alpha$, $b\mapsto\beta$ defines a homomorphism, since $\alpha$ and $\beta$ satisfy the corresponding relations ($\alpha^8=\beta^8=1, \alpha^\beta=\alpha^{-1}, \beta^\alpha= \beta^{-1}$), and this is a group of order $8$ (in fact, the quaternion group). It follows that $G$ has order at least $8$.

It seems pretty clear from the defining relations that all the elements of the group can be written in the form $a^mb^n$ with $0\leq m,n\leq 7$. But we also have the deduced relation $a^2 = b^{-2}$ and, using this, we can restrict the elements to those of the form $a^mb^n$ with $m\in\{0,1\}$ and $0\leq n\leq 3$, giving a total of eight elements.

(EDIT: As @verret rightly points out in a comment we also need to deduce, as he/she did, that $b^4 = 1$ to provide a complete argument in the paragraph above.)

It follows that the homomorphism is in fact an isomorphism, so this group is isomorphic to the quaternion group of order $8$.

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  • $\begingroup$ Ok for the first part, but why there are at most 16 elements? From the fact that every element is of the form $a^nb^m$ there are at most $8 \times 8 = 64$ elements. $\endgroup$ Commented Feb 15, 2018 at 23:58
  • $\begingroup$ @LucaSabatini Well, you had the form $a^nb^n$ (i.e., $m=n$), but I don't think that is quite correct. Let me edit this answer with a correct form. $\endgroup$ Commented Feb 16, 2018 at 0:09
  • $\begingroup$ Sorry, I'm new here. How can I let you do it? $\endgroup$ Commented Feb 16, 2018 at 0:15
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    $\begingroup$ @LucaSabatini I just meant give me a few minutes; it's done now. $\endgroup$ Commented Feb 16, 2018 at 0:20
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    $\begingroup$ @verret You are right. I didn't explain how I got that $a$ and $b$ had order (at most) $4$. I'll edit it rather than deleting though, since it shows another useful way to get the other half (showing there are no further collapses). $\endgroup$ Commented Feb 16, 2018 at 0:39

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