Let $G$ be a nonabelian group of order $8$. Necessarily$^\dagger$, $|Z(G)|=2$. Accordingly$^{\dagger\dagger}$, every noncentral elements of $G$ must have centralizer of order $4$. Therefore (orbit-stabilizer), $G$ splits into two central singleton classes, say $\mathcal C_1=\{e\}$ and $\mathcal C_2=\{x\}$, and three noncentral doublet classes, say $\mathcal D_1=\{y_1,y_2\}$, $\mathcal D_2=\{y_3,y_4\}$, $\mathcal D_3=\{y_5,y_6\}$. Since $x$ commutes with every element of $G$, then $x^k$ will do, for every integer $k$, and hence $\langle x\rangle \le Z(G)$, namely $o(x)=2$. The elements in the same conjugacy class have the same order, and at least one amongst the $\mathcal D_i$'s has its elements of order $4$, say wlog $\mathcal D_1$. So, in principle we might have both $\mathcal D_2$ and $\mathcal D_3$ made of elements of order $2$ or of order $4$, but also one-and-one. I'm looking for an argument by contradiction ruling out this latter option, as I know that the only nonabelian groups of order $8$ are $D_4$ (with two elements of order $4$) and $Q_8$ (with six elements of order $4$). Some other fact that could possibly be relevant here are:
- in general, automorphisms of a group send conjugacy classes to conjugacy classes, while preserving elements' order and classes' sizes;
- for $G$ a nonabelian group of order $8$, we get $\operatorname{Inn}(G)\cong C_2^2$ (by the "cyclic $G/Z(G)$ argument");
- there are $1\pmod 2$ subgroups of order $4$, but this is still compatible with two cylic subgroups of order $4$ (and hence four elements of order $4$) and a third subgroup of order $4$ isomorphic to Klein's 4-group (which would account for the three elements of order $2$).
So, how to rule out that $G$ has exactly four elements of order $4$ (or, equivalently, exactly three elements of order $2$)?
$^\dagger$In fact, $|Z(G)|=4$ is ruled out, as otherwise $G/Z(G)$ would be cyclic. And $|Z(G)|=1$ is ruled out by the class equation and the fact that there aren't conjugacy classes of sizes $3$ or $7$.
$^{\dagger\dagger}$For $x\in G\setminus Z(G)$, strictly: $Z(G)<C_G(x)<G$. Then apply Lagrange's theorem.
Best Answer
Suppose $y_i$ has order $4$. Then the centralizer of $y_i$ must be isomorphic to $\mathbb{Z}/4\mathbb{Z}$ and contains $x$ as an element of order $2$ and $y_i$ as element of order $4$: then $y_i^2=x$.
The case that you want to rule out is $y_i^2=x$ for $i \leq 4$, and $y_5^2=y_6^2=e$. So assume we’re in this case.
Note that $y_5x$ has order $2$ (dividing $2$ and not $1$) and is distinct from $x,y_5$, hence $y_6=y_5x$.
The subgroup generated by $y_1$ has cardinality $4$, hence index $2$, so that $y_2=y_1^{-1}=y_1x$.
Now, $y_5$ is not contained in the centralizer of $y_1$, so $y_5y_1y_5^{-1}=y_1^{-1}$, hence $(y_5y_1)^2=e$. But $y_5y_1$ is neither $e$, nor $x=y_1^2$, nor $y_5$, nor $y_6=y_5x$. Hence the contradiction.