Finite groups with only one $p$-block

Question

If $G$ is a finite group with a prime $p \big| |G|$, and $G$ has exactly one $p$-block, namely the principal block, can anything be said about the structure of $G$? I am aware that when $G$ has exactly one irreducible ordinary character in the principal block, this forces $|G|$ to be indivisible by $p$, so this question seems to be in some sense the opposite of this situation. I cannot seem to find much information about this though.

Answer 1

This is really a supplement to @DaveBenson's answer, but M.E. Harris, in Theorem 1 of his 1984 Journal of Algebra paper "On the $p$-deficiency class of a finite group", proved a rather precise theorem about a general finite group with only one $p$-block, which may be rephrased as follows: if $p$ is odd, then the finite group $G$ has only one $p$-block if and only if $O^{p}(F^{\ast}(G)) = 1$ (equivalently, $F^{\ast}(G)$ is a $p$-group). If $p = 2$, then the finite group $G$ has only one $2$-block if and only if $O_{2^{\prime}}(G) = 1$ and all components of $G$ ( if it has any components at all ) are isomorphic to $M_{22}$ or $M_{24}.$

Answer 2

If $G$ has a normal $p$-subgroup containing its centraliser then it only has one $p$-block. This goes back to Brauer.

If $G$ is simple and $p$ is odd, then $G$ has more than one $p$-block (Brockhaus and Michler, J. Algebra 1985). For $p=2$, I think the only examples with just one $2$-block are the sporadic groups $M_{22}$ and $M_{24}$, but I could be wrong.