10
$\begingroup$

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)

$\endgroup$
1
$\begingroup$

Let $H$ be a group. If $H \cong G/Z(G)$, for some group $G$, then $H$ is called a capable group. These groups have been studied y many authors and have interesting properties. For example $H$ is a nilpotent group if and only if $G$ is also nilpotent. Moreover, $c(G) = c(H) + 1$. This observation helps you to put aside some cases. For example if your group of order $48$ is nilpotent of class 2, then $H$ must be abelian (of course non-cyclic).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.