Let $G$ be a non-cyclic group of order $8$ having exactly one element of order $2$. Prove that $G$ is generated by elements $a$ and $b$ subject to the relations $a^4=1$ and $a^2=b^2$.
I can start this by a process of elimination if I know all groups of order $8$: It can not be an abelian group because it is not cyclic and all others have more then one element of order 2. It can not be the dihedral group of order $8$ $D_8=\langle x,y\mid y^4=x^2=1, xyx^{-1}=y^{-1}\rangle$ because it also has more then one element of order $2$ ($y^2$ and $x$). The only remaining group are the quaternions $H=\langle x,y\mid x^4=1, x^2=y^2, yxy^{-1}=x^{-1}$. How do I see that the third relation holds?
And is there any chance to solve this without knowing all groups of order $8$?
Best Answer
As it is noncyclic and has one element of order $1$ and one of order $2$, it has six elements of order $4$. Let $a$ be one of them, then $a^2$ is the element of order $2$. Let $b$ be outside $\{e,a,a^2,a^3\}$. Then $b$ has order $4$ and $b^2$ is the element of order $2$, that is $b^2=a^2$.