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I am hoping to solve the following exercise without having to look at all groups of order 16.

Show that the group with presentation

$$\langle a, b \,\left|\, a^4 = b^4 = 1, bab^{-1} = a^{-1} \right.\rangle $$

has order 16.


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Looking or not at tables of groups of order 16 has nothing to do with what this exercise is about by the way. – the_fox Feb 26 '12 at 18:22
Since $ba = a^{-1}b$ all elements can be put in the form $a^nb^m$ with $n,m \in \{0,1,2,3\}$ so at least $|G| \leq 16$. – WimC Feb 26 '12 at 18:36
up vote 1 down vote accepted

On the one hand, the group clearly has order less than or equal to $16$. On the other hand, note that since $bab^{-1} = a^{-1}$, we see that $ba^2b^{-1} = a^{-2}$ and $ba^3b^{-1} = a^{-3}$. Similarly, $b^2 a b^{-2} = ba^{-1}b^{-1} = (bab^{-1})^{-1} = a$ and such. So conjugating $A = \langle a \rangle$ by an element of the form $a^n$ or $b^n$ fixes $A$.

We also know that every element can be written as $a^j b^k$ for some $j,k$. It's pretty easy to see that conjugating by these elements fixes $A$ as well now. Thus $A$ is normal.

Quotient out, and see that the group has order $16$.

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Oh - I am so frequently 2ndFGITW. So it goes. – mixedmath Feb 26 '12 at 18:56

Note that the subgroup generated by $a$ is normal. It has order $4$. The quotient is given by the presentation $\langle b\ |\ b^4=1\rangle$, which also clearly has order $4$.

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