I am stuck on the following question and am looking for help/ hints:
Question: Show that if the conjugacy classes of a finite group $G$ have size at most 4, then $G$ is solvable.
(I have not been able to find anything on stackexchange that is the same or a similar question).
Context: This is a past paper question for an algebra preliminary exam I am studying for, not a homework question. Apologies if this is still against the rules. The sequence associated with the exam uses Dummit-Foote as the main textbook to give some idea of the level the questions are aimed at: I have not taken the sequence and did not learn solvable groups (or composition/ normal series) in undergrad, but have read the relevant parts of Dummit-Foote and Aluffi's Chapter 0 and tried some problems from there (I much prefer the style of exposition in the latter). So I'm quite new to the topic of solvable groups and not at all comfortable/ proficient with it yet.
My thoughts/ attempt: I have mostly been trying to work with the derived series definition, as I didn't get very far with trying to come up with an abelian/ cyclic series for $G$. I've showed that conjugacy classes are identified in the abelianization, but this just gives me that $|G/[G,G]| \leq$ (number of conjugacy classes of $G$), but the hypothesis of the question just implies that the number of conjucagy classes is at least $|G|/4$, which is unhelpful. Abelian groups are solvable but they are their abelianization, so the only a priori bound on the number of conjugacy classes is $|G|$, so I don't think this approach has gotten me anywhere.
I am mainly stuck on firstly which characterisation of solvable groups to use, and secondly how to use the bound given in the hypothesis of the question. I also noted that the hypothesis can be restated as: $ [ G \colon C_{G}(g) ] \leq 4 $ for any $ g \in G $. But $C_{G}(g)$ is not a normal subgroup a priori so I can't quotient by it (it would be if its index were bounded by 2, however).
Apologies for the length, this is my first post and I wanted to provide context for my level and the level the question is aimed at, as well as showing that I've thought and attempted the question (but am quite lost at the moment). I tried to follow the guidelines for asking questions as much as possible.
For questions on solvable groups I am feeling a bit overwhelmed by all the different approaches one can take: derived series terminating in the identity, showing $G$ has a cyclic (or merely abelian) series, finding a solvable normal subgroup $N$ such that $G/N$ is solvable... and I always find it hard to see which approach will be most viable when starting out on a problem.
At this stage I would appreciate hints etc. to point me in the right direction and get me unstuck, and not a full answer, as working out some of it on my own would be more beneficial to me.
EDIT: Thanks to the helpful hint from @DerekHolt I've managed to work out the solution so will now accept answers (if none come I will type one up when I have time) so that this can be marked as answered.
(No need for $G$ to be finite).
Let $G$ be an arbitrary group for the moment. Let $C$ be the set of conjugacy classes of $G$. Then the action on $G$ by conjugation induces a homomorphism $G\to H:=\prod_{c\in C}\mathrm{Sym}(C)$. The kernel of this action is the center of $G$.
Now suppose that every conjugacy class has cardinal $\le 4$. Then $H$ embeds into a power of $S_4$ and hence is 3-step-solvable. So $G$ itself it central-by-(3-step-solvable), hence 4-step solvable.
($G$ is said to be $k$-step-solvable if $G^{(k)}=\{1_G\}$, where $G^{(0)}=G$ and $G^{(i+1)}$ is the derived subgroup of $G^{(i)}$.)
[Note: $S_4\times S_4$ has conjugacy classes of cardinal $>4$. So there should be a (much) stronger conclusion than just being a central extension of a residually-$S_4$ group.]