A group $G$ is called perfect iff $G’ = G$.

A finite group $G$ is called immaculate iff its order is equal to the sum of orders of its proper normal subgroups.

Does there exist a finite group $G$, that is both perfect and immaculate at the same time?

Motivation behind this question:

If I am not mistaken, the notion of «immaculate groups» first appeared in this MO question by @TomLeinster. In the first paragraph of the question they said, that considered calling such groups «perfect» (because it is a group theoretic analogue of perfect numbers), however the term "perfect groups" was already taken. However, it would be interesting to know whether some of the immaculate groups are indeed perfect…

  • $\begingroup$ You mean "proper normal subgroups". These are now known as Leinster groups. Without the "normal" assumption I suspect that only cyclic groups of perfect order can be "weakly immaculate". There's just too many subgroups in non-Hamiltonian case. And Hamiltionian case can probably be reduced to abelian (which then can be easily reduced to cyclic). $\endgroup$ – freakish Jun 17 at 10:09
  • $\begingroup$ @freakish, yes I meant "normal". Thank you for pointing out my typo. $\endgroup$ – Yanior Weg Jun 17 at 12:27
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    $\begingroup$ Note that this question was already asked in a comment in the MO question you linked. It seems to still be open. (It is also claimed there are none in the GAP perfect group library, so it is probably not easy.) $\endgroup$ – verret Jun 17 at 19:56

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