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Can one classify all finite groups $G$ with the property that every proper subgroup $H < G$ is the centralizer of some $g\in G$?

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2 Answers 2

If every proper subgroup is $C_G(g)$ for some $g$ then every proper subgroup is cyclic. Indeed, $\lt g\gt$ is a proper subgroup, so $C_G(g)=\langle g\rangle$. Such groups have been described (see, for example, this description of just-non-Abelian groups. The groups must be cyclic-by-cyclic, and it is easy to decide which exactly extensions you should consider.

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As Mark mentioned above, in such a group, every subgroup must be cyclic. Indeed, it is possible to show that the only such groups are either the finite abelian simple groups (all of whom clearly satisfy the required conditions), or groups of order $pq$, where $p$ and $q$ are distinct primes (Assume that $p\gneq q$. Then, given a subgroup of order $p$ or order $q$, we have that it must be the centralizer of any element of any element in it, as if it were anything larger, we would have that our group is abelian.)

To go in the other direction, let $G$ satisfy the required conditions. If $Z(G)\neq 1$, then we have that $Z(G)=C_{G}(x)$ for some $x\in G$. Then, $x\in Z(G)$, which implies that $C_{G}(x) = G$. Thus, $G$ is abelian, and as the centralizer of any element is now $G$, we have that $G$ is cyclic of prime order.

Now note that if $H$ is any proper subgroup, then $H=C_{G}(x)$ implies that $x\in H$, and that $Z(H)\neq 1$. Also note that the assumptions on our group descend to any subgroup. Thus every subgroup of $G$ is cyclic of prime order. Let $|G|=p_{1}...p_{n}$, with $p_{i}\lneq p_{i+1}$. All the sylows of $G$ are cyclic, and the transfer map into the one of smallest order, $p_{1}$, must be surjective, giving us a subgroup of order $p_{2}...p_{n}$. As all proper subgroups of $G$ are cyclic of prime order, we have that $|G|=pq$ (I'm sure there is an easier way to show the last bit).

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