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.
Thanks.
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.
Thanks.
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 $16$.
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 $4$.