How can you show there are only 2 nonabelian groups of order 8?

It's often said that there are only two nonabelian groups of order 8 up to isomorphism, one is the quaternion group, the other given by the relations $a^4=1$, $b^2=1$ and $bab^{-1}=a^3$.

I've never understood why these are the only two. Is there a reference or proof walkthrough on how to show any nonabelian group of order 8 is isomorphic to one of these?

Solution 1:

A very down-to-Earth approach might be:

Let $G$ be a group of order 8.

Exercise 1: Show that the maximal order $m$ of an element $x$ of $G$ is either 2, 4, or 8.

Exercise 2: Show that if $m=2$, then the group $G$ is abelian.

Exercise 3: Show that if $m=8$, then the group $G$ is abelian.

Ok, so that leaves us with the case $m=4$. Let $x$ be an element of order 4. Let $H\simeq C_4$ be the subgroup generated by $x$.

Exercise 4: Show that $H$ is a normal subgroup of $G$.

Let $y\in G, y\notin H$ be a fixed element.

Exercise 5: Show that $y^2\in H$.

Exercise 6: Show that $yxy^{-1}$ is an element of order 4 in $H$.

Exercise 7: Show that either $yxy^{-1}=x$ or $yxy^{-1}=x^3$.

Exercise 8: Show that if $yxy^{-1}=x$, then $G$ is abelian.

Ok, so we must have $yxy^{-1}=x^3.$ Assume that this is the case in what follows.

Exercise 9: Show that if $y$ is of order 2, then $G$ is isomorphic to a dihedral group.

Exercise 10: Show that if $y$ is of order 4, then $y^2=x^2$.

Exercise 11: Show that if $y$ is of order 4, then $G$ is isomorphic to the quaternion group (or more precisely: the group of units of the Lipschitz' order)

Rejoice!

Remark: You won't need the result of Exercise 5 until the two last ones. I just added it there, because the element $y$ was introduced at that point.

Solution 2:

See www.math.uconn.edu/~kconrad/blurbs/grouptheory/groupsp3.pdf, which discusses groups of order p^3 for any prime p and treats the case p = 2 first.

Solution 3:

Here's the proof that there are exactly five nonisomorphic groups of order $p^3$ for every prime $p$, as it appears in Marshall Hall's Theory of Groups.

1. The abelian case is easy: you have $C_{p^3}$, $C_{p^2}\times C_p$, and $C_{p}\times C_p\times C_p$.

2. The nonabelian case. There can be no element of order $p^3$, because then the group is cyclic.

   • If all elements are of order $p$, then $p$ must be odd (otherwise the group is abelian). The center of $G$ is of order $p$ (since the quotient must be of order $p^2$ and isomorphic to $C_p\times C_p$); let $x$ and $y$ be elements of $G$ whose images generate the two cyclic factors of $G/Z(G)$. Then $x^p = y^p = 1$, and $x^{-1}y^{-1}xy\neq 1$ (otherwise, $G$ would be abelian), but must be central; so $z=x^{-1}y^{-1}xy$ generates $Z(G)$. So $G$ is given by $$G = \langle x,y,z\mid x^p=y^p=z^p=1, xy=yxz,\ xz=zx,\ yz=zy\rangle.$$

   • If there is an element $x$ of order $p^2$, then $\langle x\rangle$ is a maximal abelian subgroup of $G$, and normal (since its index is the smallest prime that divides $|G|$. Let $b\notin A$. Then $b^p\in A$, and $b^{-1}ab=a^r$ for some $r$; since $G$ is nonabelian, $r\neq 1$. Since $b^p$ commutes with $a$, $a^{r^p}=a$, so $r^p\equiv 1\pmod{p^2}$. From Fermat's Little Theorem, $r^p\equiv r\pmod{p}$, so $r\equiv 1\pmod{p}$.

     Write $r=1+sp$, and let $j$ such that $js\equiv 1\pmod{p}$. Then $$b^{-j}ab^j = a^{r^j} = a^{(1+sp)^j} = a^{1+spj} =a^{1+p}.$$ Since $(j,p)=1$, $b^j\notin A$, so replacing $b$ with $b^j$, we may assume that $b^{-1}ab=a^{1+p}$.

     Now, $b^p = a^t$, and $t$ must be a multiple of $t$, because $b$ is not of order $p^3$. Write $t=up$, so $b^p=a^{up}$. Then we have: $$\begin{align*} (ba^{-u})^p &= b^pa^{-u(1+(1+p)+(1+p)^2 + \cdots + (1+p)^{p-1})}\\ &= b^p a^{-up-up(1+2+\cdots + p-1)}\\ &= b^p a^{-up-up\binom{p}{2}}. \end{align*}$$

     • If $p$ is odd, then $up\binom{p}{2}$ is a multiple of $p^2$, so we get $(ba^{-u})^p = b^pa^{-up} = b^pb^{-p} = 1$. Setting $c=ba^{-u}$ we get $c^{-1}ac = b^{-1}ab$, so the group is presented by $$\langle a,c\mid a^{p^2} = c^p = 1,\ ac = ca^{1+p}.\rangle$$

     • If $p$ is $2$, however, we get $(ba^{-u})^2 = b^2a^{-up-up} = b^2$. We have two possibilities: it could be that $b^2=1$, in which case we get the same presentation as above: $$\langle a,b\mid a^{4} = b^2 = 1,\ ab=ba^3\rangle.$$ Or it could be that $b^2=a^2$; we must have $b^{-1}ab=a^3$ (it cannot equal $a$, because then $a$ and $b$ commute and $G$ is abelian), so the group is given by $$\langle a,b\mid a^4=1,\ a^2=b^2,\ ab=ba^3.$$

Burnside uses essentially the same approach, though he only deals explicitly with odd $p$; the classification for groups of order $p^3$ takes up about two pages (one paragraph, but he invokes results covering two previous pages). He then proceeds to those of order $p^4$; that takes four and a half pages (plus invoking stuff that covers at least one previous page). He only lists those of order $2^3$ and $2^4$.