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There are more than 50 groups of order 48, and among them 16 groups have center of order 2, let $G$ be among such groups. Then $G/Z(G)$ is a group of order 24. What is this group of order 24? There are 15 groups of order 24, and among them, only four groups can occur as central quotient of group of order 48. These are $D_{24}$, $S_4$, $C_2\times C_2\times S_3$, and a group of type $(C_6\times C_2)\rtimes C_2$. (These observations are written with the help of GAP.)

So, there are large number of groups of order 24, which are not central quotients of groups of order 48.

Question What are necessary and/or sufficient conditions for a finite group $H$ to be central quotient of a finite group $G$.

(Note: Here, by central quotient, we mean quotient by the (full) center of the group. It is well known that cyclic groups can not be central quotients of any group; finite or infinite)

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This question might focus on conditions relevant to groups of order 24 (so not focusing on abelian and low nilpotency degree). – Jack Schmidt Apr 30 '13 at 3:57
I'd like to see a proof that the other 11 groups of order 24 are not inner automorphism groups. Is it really so easy to find necessary conditions? [ Rucha checked |G|=48, and I checked much higher orders for |G|, but how big can the smallest |G| be? ] – Jack Schmidt May 2 '13 at 19:12
SL(2,3) and 3 x Q8 are not, by – Jack Schmidt May 2 '13 at 19:48
Ellis et al.'s method settles the question for groups of order 24 (only the 4 found are central quotients). – Jack Schmidt May 2 '13 at 19:59

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