Let G be a group of order $2^n$ for $n\geq 2$. Show that G has at least $4$ conjugacy classes Let $G$ be a group of order $2^n$ for $n\geq 2$. 
(A) Show that $G$ has at least 4 conjugacy classes; 
(B) Show that if $G$ has exactly 4 conjugacy classes, then $n = 2$ or $n = 3$
For the case of $n=2$
Order of G is $2^2=4$
By corollary (Dummitt and Foote pg.125)
If G is a group of order $p^2$ where $p$-prime,
then $G$ is abelian.
So for the case of $n=2$,
$G$ is abelian, and so the singletons $\{a\}$ for a in $G$ are the conjugacy classes. Thus there are $4$ conjugacy classes.
This is all I have. I have no idea how to go for $n=3$ and there on.
 A: Since $G$ has a nontrivial center, it must have at least $2$ conjugacy classes. If $|Z(G)| = 4$, we're done. Suppose $Z(G) = 2$. As you note, $G$ is abelian if $|G| = 4$, so we may take $|G| \geqslant 8$, i.e. $n \geqslant 3$. By the class equation, the remaining $2^{n}-2$ elements are partitioned into nontrivial conjugacy classes. If there were only $3$ conjugacy classes total, then the remaining conjugacy class would have $2^{n}-2$ elements. But this is a contradiction, as the order of any conjugacy class must be a power of $2$, and $2^{n}-2$ is not a power of $2$ for $n \geqslant 3$. Hence, we must have at least $4$ conjugacy classes. 
Now suppose that there are exactly four conjugacy classes. As above $Z(G) \geqslant 2$; since there are exactly four conjugacy classes, $Z(G) \leqslant 4$. If $Z(G) = 4$, then $G = Z(G)$, so $G$ has order $4$. Suppose $Z(G) = 2$, and $n \geqslant 3$. Then as above, the remaining two conjugacy classes compose the remaining $2^{n}-2$ elements. Let one of the remaining two conjugacy classes have order $2^{k}$ for some $1 \leqslant k < n$. Then the other conjugacy class must have order $(2^{n}-2)-2^{k} = 2(2^{n-1}-2^{k-1}-1)$. This must be a power of $2$, but if $k > 1$, then $2^{n-1}-2^{k-1}-1$ must be odd. Hence, we must have $k = 1$, in which case $2^{n-1}-1-1 = 2^{n-1}-2$ must be a power of $2$. As above, this is only the case if $n = 3$, so the only other possible candidate is $n =3$. 
Note that this shows the only possibilities are $n = 2, n = 3$, but in fact, no group of order $8$ has exactly $4$ conjugacy classes. Both $D_{4}$ and $Q_{8}$ have $5$ distinct conjugacy classes, and the remaining groups of order $8$ (up to isomorphism) are abelian. I'm guessing there's a slicker way to make the above arguments than the sort of number theory fiddling I've done, but it does work. 
A: Assume there are only 3 conjugacy classes. Then two of them are singletons, since the center is non-trivial. The other one would have size $2^n-2$, which is clearly not a divisor of the group order $2^n$ if $n$ is at least $3$.
For the second part you should note that $2^{n-1}+2^{n-2}+2$ is strictly smaller than $2^n$ when $n$ is at least $4$. So there is no way to have only 4 conjugacy classes in that case.
