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show that the relations $a^4=1$, $b^2=a^2$, $b^{-1}ab=a^{-1}$ define a group of order 8.

This is a homework problem I got. I don't understand what they mean by 'a group of order 8', because I think it only makes sense to talk about the order of an element in a group. For example, the element 2 has order 2 in the group $Z_4$.

Should I try to show that it can be generated by an element of order 8? I don't think this is right though, because $a^4=1$, so the least order should be 4.

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This just means the group has 8 distinct elements. – Alex Oct 12 '11 at 6:51
"Group of order 8" means "a group with 8 elements". That is, the order of a group is the size of its underlying set. – Arturo Magidin Oct 12 '11 at 6:52
up vote 7 down vote accepted

It means that the group that is presented by $$\langle a,b\mid a^4 = 1, b^2=a^2, b^{-1}ab = a^{-1}\rangle$$ is a group with exactly $8$ elements. That is, the most general group that satisfies these relations has 8 elements.

I would suggest finding a normal form for the elements (perhaps they can all be written as $a^ib^j$ with $i$ and $j$ in certain range?) to give an upper bound, and then find a group you know that has $8$ elements, and which has two elements that satisfy the given relations and generate, to show that the group presented has at least $8$ elements (by von Dyck's theorem).

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