# Non cyclic group of order $8$ having exactly one element of order $2$

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 $8$?

• What are the elements of order $2$ in the quaternion group? Aug 15, 2018 at 18:38
• There are abelian groups of order 8 that are not cyclic, by the way. There are $3$ abelian groups of order 8: $\mathbb{Z}_8, \mathbb{Z}_2 \times \mathbb{Z}_4, \text{ and } \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$. Aug 15, 2018 at 19:17
• I know that and as I wrote the two which are not cyclic have more then one element of order $2$ Aug 15, 2018 at 20:33

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 $b^2=a^2$.
• I understand what you wrote but I don't see why this is all what I need. I understand that $G$ has two elements $a$ and $b$ and that they satisfy the relations. But how do you know, that you just need the generators $a$ and $b$? And how do you know that together with these 2 relations you get a group of order 8? I have difficulties to simplify expressions. We have elements $e,a,a^2=b^2,a^3,b,b^3=ba^2=a^2b,ab,ba,aba,ab^3=a^3b...$ Some of them must be equal but how do I see that? Aug 15, 2018 at 20:51
• And how would an isomorphism to $D_8$ with the presentation I gave above would look like? Aug 15, 2018 at 20:55
• $a$ generates a subgroup of order $4$. Since $b$ is not in that subgroup, $a$ and $b$ must generate a larger subgroup, but the only possible size is $8$ ie, the whole group. Also, if you haven't gathered from the hints, $D_8$ has multiple elements of order $2$ and it is actually the Quaternion group you are looking for. Aug 15, 2018 at 23:00
• @mathstackuser Observe that $bab^{-1}$ has order $4$, and $bab^{-1}$ cannot be a power of $b$ (why?). So it must be a power of $a$ of order $4$. This shows that $bab^{-1}=a^{-1}$. ($ab$ must be of order $4$, so $abab^3=e$.) Aug 16, 2018 at 12:56
• Because $(bab^{-1})^4=ba^4b^{-1}=e$ and $(bab^{-1})^2=bb^2b^{-1}=b^2\ne e$. The argument in the parenthesis means $a\circ bab^{-1}$ is identity, so $bab^{-1}$ is $a^{-1}$ by definition in fact. By the way, that $ab$ is of order $4$ can be seen from $ab\ne a^2$ and $ab\ne e$. Aug 16, 2018 at 13:09