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Let $ G $ be a perfect group. We often consider extensions $$ 1 \to A \to E \to G \to 1, $$

where $ A $ is abelian, $ G $ is perfect and the image of $ A $ is central. Is it possible for such an extension not be central? In other words if we have $$ 1 \to A \to E \to G \to 1, $$ where $ A $ is finite abelian, $ G,E $ finite perfect then must the image of $ A $ be central?

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2 Answers 2

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Let $W := \mathbb{Z}_2 \wr A_5$ and let $G := \operatorname{Inn}(W)$. This is a perfect group of order $960$, has trivial centre, and is of the form $(\mathbb{Z}_2)^4 \rtimes A_5$.

The following GAP code can identiy more examples:

n := 5000;
for i in [ 1..n ] do
    for j in [ 1..NrPerfectGroups( i ) ] do
        G := PerfectGroup( i, j );
        Ns := Filtered( NormalSubgroups( G ), N -> IsAbelian( N ) and not IsSubgroup( Centre( G ), N ) );
        if not IsEmpty( Ns ) then
            Print( "PerfectGroup( ", i, ", ", j, " ) is an example.\n" );
        fi;
    od;
od;
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Let $A = \left\{ x \in (\mathbb{Z}/2)^5 : \sum_i x_i = 0 \right\}$, and let $G = A_5$ act on $A$ by permuting coordinates. Then $$ 0 \to A \to G \ltimes A \to G \to 1 $$ is a non-central extension you are looking for.

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  • $\begingroup$ I appreciate the especially simple construction here, I mostly accepted the other answer because of the GAP code. Does your construction use some fact along the lines of: for any irreducible representation $ \pi$ of a perfect group $ G $ on a finite vector space $ A $ then $ A \rtimes_\pi G $ is perfect? $\endgroup$ Apr 28, 2022 at 1:56
  • $\begingroup$ I don't think so, since there seems to be no reason why $[G,A]$ should be a sub vector space of $A$. This is just the canonical example in my opinion: $A_5$ is the smallest perfect group, and $E = A_5 \ltimes (\mathbb{Z}/2)^5$ is the most simple non-trivial extension of $A_5$ I can think of. Unfortunately, $E$ is not perfect, but its derived subgroup is. $\endgroup$
    – Dune
    Apr 28, 2022 at 13:46

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