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A group is subdirectly irreducible provided it has a least nontrivial normal subgroup. Subdirectly irreducible groups are also referred to as monolithic groups in the literature. Every simple group is sub-directly irreducible, but there are many subdirectly irreducible groups that are not simple.

Is there any classification for subdirectly irreducible groups?

Any comments or hints are highly appreciated.

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  • $\begingroup$ I don't understand your first sentence. It seems to me that a group is subdirectly irreducible if the intersection of any two nontrivial (i.e., $\neq\{1\}$) irreducible subgroups is nontrivial. $\endgroup$
    – YCor
    Sep 23, 2014 at 11:50
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    $\begingroup$ I am assuming that it means that the group has a unique minimal normal subgroup. If we restrict attention to finite groups, then any irreducible representation of any group over a prime field gives rise to a semidirect product with this property. Similarly, for any finite nonabelian simple group $S$ and any transitive permutation group $P$, $S \wr P$ has the property. So there are far too many examples for there to be a complete classification. So I think you need to ask a more directed question. $\endgroup$
    – Derek Holt
    Sep 23, 2014 at 11:55
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    $\begingroup$ @YCor, in universal algebra subdirectly irreducible usually means not a subdirect product of an arbitrary collection of proper quotients. This is equivalent to a unique minimal nontrivial normal subgroup. You are thinking of not being a subdirect product of two proper quotients, which is another reasonable interpretation $\endgroup$ Sep 23, 2014 at 13:18

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I don't think there is a classification of finite monolithic groups. Here is an extended comment. I am assuming the group $G$ is finite. The minimal normal subgroup has no characteristic subgroup and hence is isomorphic to $T^n$ where $T$ is a finite simple group. If $T$ is cyclic of prime order then the action by conjugation of $G$ on $N$ must be an irreducible representation, as Derek's comment indicates, since an invariant subspace would give a smaller normal subgroup.

If $T$ is non-abelian the action of $G$ on $N$ by conjugation is faithful because the centralizer of $N$ is normal and hence contains $N$ or is trivial.

If you assume that no proper subgroup of $G$ generates the same variety as $G$ then more can be said and you might consult the book of Hanna Neumann, Varieties of Groups.

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    $\begingroup$ You can say a little bit more. If $T$ is nonabelian, then $N$ is self-centralizing, and $T \wr P \le G \le {\rm Aut}(T) \wr P$ for some transitive $P \le S_n$. If $T$ is abelian, then I think $C_G(N)$ has to be a homocycl;ic $p$-group $C_{p^k}^n$ for some prime $p$ and $k>0$. $\endgroup$
    – Derek Holt
    Sep 23, 2014 at 13:59
  • $\begingroup$ Sorry, I withdraw that last claim. $C_G(N)$ need not be abelian. For example ${\rm GL}_2(3)$ has a unique minimal normal subgroup, which has order $2$, and is central. $\endgroup$
    – Derek Holt
    Sep 23, 2014 at 19:01

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