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If $G$ is a group such that every non-trivial subgroup is isomorphic to $G$ itself, then $G= \mathbb{Z}$ is the only infinite group with that property (up to isomorphism). Amongst the finite groups we essentially have $\mathbb{Z}/p\mathbb{Z}$ for $p$ prime.

Turning arrows around, let us say that a group $G$ has property $\text{Q}$ if

whenever $H$ is a non-trivial group and $s:G\to H$ is a surjective group homomorphism, then $H\cong G$.

Again $\mathbb{Z}/p\mathbb{Z}$ has this property for $p$ prime. (I don't know whether they are the only finite groups with property $\text{Q}$.) Note that $\mathbb{Z}$ does not have property $\text{Q}$.

Question. For which cardinals $\kappa\geq\aleph_0$, if any, is there a non-simple group with property $\text{Q}$ having cardinality $\kappa$?

Acknowledgement. Thanks to Dave Benson for suggesting to consider non-simple groups only.

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    $\begingroup$ @HJRW This is not true. For example, Prüfer groups $\mathbb Z(p^\infty)$ have property Q even though they are not simple. Note that the definition does not require $s$ itself to be an isomorphism. $\endgroup$ Commented Nov 28, 2023 at 9:13
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    $\begingroup$ Here is a Math.SE question dealing with the non-simple groups in $Q$. (Take home points are: finite generation implies simple, and there are non-abelian countable examples.) Of course, simplicity is enough to answer the stated question (e.g.), so I agree with @HJRW re closure. $\endgroup$
    – ADL
    Commented Nov 28, 2023 at 9:21
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    $\begingroup$ By general universal algebra, any group is a subdirect product of subdirectly irreducible groups that are its quotients, thus every group with property Q must be subdirectly irreducible. In particular, this easily implies that the only abelian groups with property Q are $C_p$ and $\mathbb Z(p^\infty)$. $\endgroup$ Commented Nov 28, 2023 at 9:21
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    $\begingroup$ If the question were to understand the non-simple groups with property $Q$, I think it might not get closed. $\endgroup$ Commented Nov 28, 2023 at 9:24
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    $\begingroup$ @HJRW The definition is formulated precisely, and does not require $s$ to be an isomorphism. This is clearly intended, as the motivating dual property in the first paragraph likewise does not require the inclusion map itself to be an isomorphism, lest $\mathbb Z$ would not have the property. I am perfectly aware of the OP’s history of non-research-level questions, but that does not warrant arbitrarily changing the meaning of his question to make it more trivial than it is. $\endgroup$ Commented Nov 28, 2023 at 11:10

2 Answers 2

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CW answer.

As already mentioned in a comment, the question was already asked at MathSE. The property in consideration is "every nontrivial quotient is isomorphic to the group itself".

In summary:

  • the trivial group satisfies this property
  • simple groups satisfy this property
  • as mentioned in this answer, finitely generated groups with this property are trivial or simple (indeed, every nontrivial f.g. group has a simple quotient). This more generally holds for groups that are finitely normally generated.
  • as mentioned in the same answer the most obvious nontrivial nonsimple examples are Prüfer (aka quasi-cyclic) groups $C_{p^\infty}$ for $p$ prime. These are the only ones among abelian groups.
  • nonabelian nonsimple examples are provided in this answer. (As the Prüfer groups, they are locally finite; these ones are direct limits of iterated wreath products of finite alternating groups.)
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    $\begingroup$ Thanks for taking seriously my suggestion for reinterpreting the question. The question of understanding the non-abelian non-simple groups with the given property seems like an interesting one that has not really been satisfactorily answered at this point. $\endgroup$ Commented Nov 28, 2023 at 22:06
  • $\begingroup$ @DaveBenson I don't think I'm "reinterpreting". The question is quite clear, stated twice (in title and body). The meaning of "isomorphic" is clear-cut: "there exists an isomorphism". When somebody uses "is isomorphic" to mean "some canonical map is an isomorphic" it's just ill-stated. I saw little point in using this wrong interpretation to get an uninteresting question. $\endgroup$
    – YCor
    Commented Nov 29, 2023 at 6:21
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    $\begingroup$ That's not what I meant. I was referring to my comment below the question. $\endgroup$ Commented Nov 29, 2023 at 8:37
  • $\begingroup$ @DaveBenson I also understood you're referring to your comment "If the question were to understand the non-simple groups with property Q, I think it might not get closed." My point is that first I'm not sure I noticed this comment, and also I didn't think of my answer as a "reinterpretation". However, I didn't address the "large cardinal" question in this cw answer, and indeed in this case, as you suggested, the question needs to be modified to ask about only about nonsimple pseudosimple groups. $\endgroup$
    – YCor
    Commented Nov 29, 2023 at 10:35
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References & terminology.

An algebraic structure is called pseudosimple if it has more than one element and is isomorphic to all its nontrivial quotients. This concept was introduced in the paper

Monk, Donald
On pseudo-simple universal algebras.
Proc. Amer. Math. Soc. 13 (1962), 543-546.

The main result of the paper is that if $A$ is pseudosimple algebra, then the congruence lattice of $A$ is a well-ordered chain of order type $\omega^{\beta}+1$ for some ordinal $\beta$, and, conversely, for every ordinal $\beta$ there exists a pseudosimple lattice whose congruence lattice is a well-ordered chain of order type $\omega^{\beta}+1$.

A later paper that is closer to the problem on this page is:

Schein, Boris M.
Pseudosimple commutative semigroups.
Monatsh. Math. 91 (1981), no. 1, 77-78.

The main result here is that a commutative semigroup is pseudosimple iff it is a $2$-element commutative semigroup, $\mathbb Z_p$, or $\mathbb Z_{p^{\infty}}$ for some prime $p$. This paper states as a corollary to the main result that an abelian group is pseudosimple iff it is isomorphic to $\mathbb Z_p$, or $\mathbb Z_{p^{\infty}}$

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    $\begingroup$ The result on the congruence lattice is very nice! $\endgroup$
    – YCor
    Commented Nov 28, 2023 at 17:32
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    $\begingroup$ The result on abelian groups (which I mentioned in my post, although not aware that it's stated somewhere) is very easy because every nontrivial abelian group either has a finite quotient, or else is divisible and hence has a Prüfer quotient. A similar conclusion holds for modules over any fixed PID $\endgroup$
    – YCor
    Commented Nov 28, 2023 at 17:33

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