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A subgroup $H$ of a group $G$ is called retract of $G$ if there exists homomorphism $r:G\longrightarrow H$ so that $r\circ i=id_H$, where $i:H\hookrightarrow‎‎ G$ denotes the inclusion map. Also, recall that a group ‎$‎G‎$‎ is Hopfian if every epimorphism ‎$‎f :G‎\longrightarrow ‎G‎‎$‎ is an automorphism.

''Is retract of a finitely generated Hopfian group Hopfian?''

Thanks in advance.

Shaun
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    How about a (restricted) wreath product of $A_5$ with your favourite nonHopfian group, such as the Baumslag-Solitar group ${\rm BS}(2,3)$? I think all nontrivial normal subgroups contain the base group of the wreath product. ButI guess you still have to prove that this is not a quotient of ${\rm BS}(2,3)$. – Derek Holt Sep 30 '17 at 09:02
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    On second thoughts, replace $A_5$ by an uncountable simple group. – Derek Holt Sep 30 '17 at 09:21
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    @DerekHolt If you're very lucky, then your favourite nonHopfian group is an infinite direct sum of cyclic groups of order two, rather than $BS(2,3)$. Then it's rather easier to prove that something is not a quotient of it! – Jeremy Rickard Sep 30 '17 at 12:30
  • If $G$ is Hopfian and abelian, then every retract $H$ of $G$ is also Hopfian. For this, assume that $f:H\longrightarrow H$ is an epimorphism. Since $G$ is abelian. then $H$ is a direct summand of $G$. So there is subgroup $K$ so that $G=H\oplus K$. Now define $\bar{f}:G\longrightarrow G$ by $\bar{f}(hk)=f(h)k$. Clearly, $\bar{f}$ is an epimorphism. Since $G$ is Hopfian, then $\bar{f}$ (and so $f$) is an isomorphism. It follows that $H$ is Hopfian. –  Sep 30 '17 at 16:06
  • In general, if $H$ is a retract of $G$ with retraction $r:G\longrightarrow H$, then we have $G=H\ltimes Ker(r)$. If $f:H\longrightarrow H$ is an epimorphism, then I can define $\bar{f}:G\longrightarrow G$ by $\bar{f}(hx)=f(h)x$ for $h\in H$ and $x\in Ker(r)$. But I can't prove that $\bar{f}$ is a homomorphism:( –  Sep 30 '17 at 16:14
  • @JeremyRickard I guess I was looking for a finitely generated example, and I still have not found one! – Derek Holt Oct 01 '17 at 13:22
  • @DerekHolt Do you know that your first suggestion (restricted wreath product of $A_5$ with $BS(2,3)$) doesn't work? – Jeremy Rickard Oct 01 '17 at 17:22
  • @JeremyRickard No, but I think it might be hard to prove that it does work, because you would have to consider all surjective homomorphisms $G \to G$, and then look at the images of their kernels. – Derek Holt Oct 01 '17 at 18:27

1 Answers1

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An adaptation of Derek's suggestion works.

Let $I$ be a set, $S$ a non-abelian simple group, $S^{(I)}$ the restricted power.

Proposition 1: every normal subgroup of $S^{(I)}$ equals $S^{(J)}$ for some $J\subset I$.

(Proof: standard exercise)

Corollary 2: for every group $G$, the only normal subgroup of $S\wr G=S^{(G)}\rtimes G$ contained in $S^{(G)}$ are $\{1\}$ and $S^{(G)}$.

(Proof: by the proposition it should be $S^{(J)}$ for some $J\subset G$, and from the normality assumption $J$ has to be left $G$-invariant.)

Corollary 3: now fix $S$ finite. let $G$ be a group with no nontrivial normal subgroup of exponent dividing the exponent of $S$, and such that $S\wr G$ is not isomorphic to a quotient of $G$. Then $S\wr G$ is Hopfian.

(Proof: since $G$ acts faithfully on $S^{(G)}$, every nontrivial normal subgroup intersects nontrivially $S^{(G)}$, and hence any proper quotient of $S\wr G$ is a quotient of $G$, so the second assumption ensures the conclusion.)

It remains to get an example. Fix $S$ finite nonabelian simple group. Fix $p$ prime not dividing $|S|$. Consider Ph. Hall's group $G_p$, defined as the quotient of the group of matrices $$m(n,x,y,z)=\begin{pmatrix}1&x&z\\0&p^n&y\\0&0&1\end{pmatrix},\quad (n,x,y,z)\in\mathbf{Z}\times\mathbf{Z}[1/p]^3$$ by the central cyclic subgroup generated by $m(0,0,0,1)$. It's well-known that $G_p$ is non-Hopfian (hint: use conjugation by the diagonal matrix $(p,1,1)$). Also, $G_p$ is finitely generated (by $\{m(1,0,0,0),m(0,1,0,0),m(0,0,1,0)\}$). Clearly its only order of torsion is given by powers of $p$, so the first assumption of Corollary 3 (on the exponent) holds. In addition, $S\wr G_p$ is isomorphic to a quotient of $G_p$, since $G_p$ is solvable but not $S\wr G_p$.

To conclude: for every prime $p$ and every finite non-abelian simple group $S$ of order coprime to $p$, $S\wr G_p$ is a Hopfian group, while its retract $G_p$ is non-Hopfian.

YCor
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  • @ YCor Your answer is great. Thank you for your exact explanation. Your example can be so useful for my other works. Thank you very much. –  Oct 12 '17 at 14:56