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I have the following structure $$\langle a,b\mid a^5=a, b^9=b, a^2=b^2, ab=b^7a\rangle$$ GAP tells me that this is of order $8$. Is it probably the quaternion group? How could GAP show it? Thanks!

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    What does "$a^5=5$" mean in this context? $\qquad$ – 2017-01-12
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    Note that proper notation is $\langle a,b\mid a^5=5, b^9=b, a^2=b^2, ab=b^7a\rangle,$ not $.\qquad$ – 2017-01-12
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    @MichaelHardy I'd guess the second $5$ should be $a$ for $a^5 = a$. But it's certainly better to wait for OP to confirm. – 2017-01-12
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    Among groups of order $8$ there are three abelian groups, $\mathbb Z_2^3$, $\mathbb Z_4\times\mathbb Z_2$, and $\mathbb Z_8,$ and two non-abelian groups: the dihedral group and the quaternion group. So you just need two things: (1) How do you show there are only those five?, and (2) Is the given information consistent with just one of those groups? However, if the statement that it is of order $8$ is wrong, then there's more to do than that. – 2017-01-12
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    @MichaelHardy: what quid stated is right. I meant $a$. I edited it as it should e. – 2017-01-12
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    Which structure are you considering? A semigroup, a monoid, a group? – 2017-01-12
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    @J.-E.Pin: I worked through some semigroups and see that this one has order $8$ and it is regular by GAP. – 2017-01-12
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    @MichaelHardy: As I am doing GAP, it helps me showing the structure is non abelian. – 2017-01-12
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    I don't know how GAP could should it, but every element is one of three forms: $b^m$, $b^ma$ or $a$ for $1\leq m \leq 9$, so it wouldn't be hard to make a multiplication table and check if it's a group. – 2017-01-12
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    @MiloBrandt: Yes that's right. it has a very finite one and in fact it has just one idempotent. If I don't get any hint, I will do it by hand so far. Thanks. – 2017-01-12
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    It's easy to see that a group given by these relations is $Q_8$. Of course, the question is how to check this in GAP. If you're dealing with a semigroup in GAP (call is `S`, for example), then call `AsGroup(S)`. If the elements of `S` form a group, it will return this group (otherwise it returns `fail`). Then you can explore that group and check whether it is $Q_8$ (see `IdGroup`, `QuaternionGroup`, `StructureDescription`). If you'll show how did you create such semigroup in GAP, I could demonstrate this in more details, using your example. – 2017-01-12

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The quaternion group with the usual notation satisfies these equalities: \begin{align} ij & = k & ji & = -k \\ jk & = i & kj & = -i \\ ki & = j & ik & = -j \\ i^2 = j^2 = k^2 & = -1 \end{align} What you have is consistent with $a=i$ and $b=j$.

It is not consistent with the dihedral group of order $8$. If two elements $a,b$ generate the dihedral group, then at least one of them must have order $2$, since all the other elements only generate the rotation group of order $4$. But they cannot both have order $2$, since then they generate a subgroup of order $4$ in which each non-unit element has order $2$. So if they're in the dihedral group, we'll have to assume one has order $4$ and the other has order $2$. But that is not consistent with $a^2=b^2$.

I'll leave it as an exercise to rule out the three abelian groups of order $8$.

Hence if this is a non-abelian group of order $8$, then it's the quaternion group.