Conjugacy classes of a group of order $8k$ Let G be a group of order $8k$, show that there are at least 5 different conjugacy classes.
Hi everyone, I have this problem I think I had a solution involving stabilizers, however I feel there must be a really easy and quick solution. Any ideas on what that might be?
Edit: To give more motivation into the question. It was suggested to me by a college, I believe he found it a book by J.M. Gamboa Mutuberría(he has a bunch, I do not know which).
My current line of thought consist of using the fact that considering the action 
$$
\rho:G\times G\rightarrow G, \rho(x,y)=yxy^{-1}
$$
and that $x\in G$ is a fixed point iff $x$ is in the centralizer of $G$ we have $$
card(G)=card(Centr(G))+\sum_{s=1}^{n}\theta_s
$$, where $\theta_1\ldots\theta_n$ are the distinct orbits, now orbits have at least size 2 and divide 8k. But I said before, I think this is getting to complicated and my college suggested the problem must have a simpler suggestion
 A: A simple answer can be based on the fact that a $2$-group has a non-trivial center (the same applies to any $p$-group). The assumption about the order $n=8k$ of the group $G$ implies by basic Sylow theory that there is a subgroup $P\le G$ of order $8$. Let $z\neq1_G$ be an element of $Z(P)$. It follows that the centralizer of $z$ has order
$8\ell$, $\ell\ge1$. Therefore the conjugacy class of $z$ has $|[z]|=n/(8\ell)\le n/8$
elements.
Assume that there are at most four conjugacy classes — two in addition to $[1]$ and $[z]$. They have sizes $n/a$ and $n/b$ for some divisors $a,b>1$ of $n$. The conjugacy classes form a partition of $G$, so
$$
1+\frac n{8\ell}+\frac n a+\frac n b=n.\qquad(*)
$$
If $a=b=2$ then the left hand side of $(*)$ is too large. If $a,b\ge3$, then the l.h.s. is too small ($1+n/8<n/3$ whenever $n\ge8$). Hence we can assume w.l.o.g. that $a=2$ and $b>2$. Also, $1+n/8\le n/4$ leading to $b\le 4$.


*

*If $b=4$, then $(*)$ implies that $\ell=1$ and $n=8$. But, a non-abelian group of order eight is either the quaternion group $Q_8$ or the dihedral group $D_4$. Neither has a conjugacy class of size four, so this case can be ruled out.

*If $b=3$, then $(*)$ implies that $\ell=1$ and $n=24$. A conjugacy class of size $n/3$ necessarily consists of elements of order three, each generating its own centralizer. Thus $G$ has four Sylow $3$-subgroups. The conjugation action on the set of those Sylow subgroups gives us a homomorphism $f:G\to S_4$. An element $x\in G$ of order three can normalize only one Sylow $3$-subgroup, namely the one it generates, so $f(x)$ is a 3-cycle. It follows that all the 3-cycles of $S_4$ are in the image of $f$, and those are known to generate $A_4$. Therefore $\operatorname{ker} f$
has order one or two. The alternative of a kernel of size two is impossible. After all, the kernel is normal, and consists of full conjugacy classes. Hence $f$ is an isomorphism, and $G\cong S_4$. But $S_4$ has five conjugacy classes. A contradiction.

A: Let $G$ be a finite group with $|G| \equiv 0$ mod $8$ and assume that the number of conjugacy classes is at most $4$. This is equivalent to the statement that the number of irreducible complex characters of $G$ is at most $4$. Since $|G|$ is the sum of the squares of its irreducible character degrees we get $|G|=1+a^2+b^2+c^2$, where the $1$ comes from the principal character and $a,b,c$ are non-negative integers. Since squares mod $8$ equal $\bar{0}$,$\bar{1}$ or $\bar{4}$, the equation $\bar{0}=\bar{1}+\bar{a}^2+\bar{b}^2+\bar{c}^2$ mod $8$ has no solutions, a contradiction.
Note (1) One can show that if $|G|$ is odd and $k(G)$ denotes the number of conjugacy classes, then $|G| \equiv k(G)$ mod $16$.
(2) In general, for a non-abelian $G$,  $k(G) \leq \frac{5}{8}|G|$. Hence if we have equality here, then $|G|$ must be a multiple of $8$. Example: $G=Q_8$, the quaternion group of order $8$.
