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Let $G$ be a group, say nontrivial. Aluffi interprets the property

"The only homomorphic images of $G$ are $1$ and $G$ [paraphrased]" ($\dagger$)

to mean

"If there exists a surjective homomorphism $\varphi \colon G \twoheadrightarrow G'$, then either $G' \cong 1$ or $G' \cong G$ [paraphrased]." $(*)$

The context is we want to show such groups $G$ satisfying $(*)$ are simple. The way to do this is to assume otherwise, take a nontrivial proper normal subgroup $N$ of $G$, and look at $\pi \colon G \twoheadrightarrow G/N$.

Question: How can we show it is impossible that $G \cong G/N$ under some other homomorphism $\varphi$? For example, we could have something like $\prod^\infty \mathbb{Z} \cong \prod^\infty \mathbb{Z} / (\mathbb{Z}, 0, 0, \dots)$. Of course, this group doesn't satisfy $(*)$ (just project onto the first coordinate), so it isn't exactly a counterexample, but could some nasty group exist that is? What prevents this?

Is it better to think of $(\dagger)$ to mean

"Every surjective homomorphism out of $G$ is either trivial or an isomorphism?" $(**)$

Is this what Aluffi actually meant by $(*)$? Does it not matter? That is, are $(*)$ and $(**)$ actually equivalent as written?

Thanks in advance.

Related: Definition of Simple Group

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Doug
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  • It's easy to prove that $G/N$ is not isomorphic to $G$ if $G$ is finite, but is not always true if $G$ is infinite, for arbitrary normal subgroup $N$, as you've noticed. – Thomas Andrews Oct 14 '14 at 05:08
  • It is probably best to read it the second way. It's certainly enough for most purposes. I would not be surprised if the first reading was true, but it is probably not needed – Thomas Andrews Oct 14 '14 at 05:15
  • The second way of course implies simplicity immediately. – Doug Oct 14 '14 at 05:21
  • Which is another reason to think that is what is meant. Depends on the book, but a book which says this, gives no proof, and doesn't leave it as an exercise would be guilty of malpractices rather than sloppiness. :) – Thomas Andrews Oct 14 '14 at 05:22
  • Well, it is, in fact, the exercise to show that (*) implies simplicity. – Doug Oct 14 '14 at 05:24
  • It would then depend on the level of the other exercises in the section. I'd still guess the second interpretation, but I'd have to see the book to really feel confident. If the other questions are barely-disguised "check that you understand the definitions"-type questions, then it almost certainly means the second. – Thomas Andrews Oct 14 '14 at 05:28
  • Right, and that's exactly the type it is. So my thought is that he wasn't being careful enough. It certainly isn't meant to be a deep exercise; 4th question from the section on Sylow Theorems. – Doug Oct 14 '14 at 05:31
  • If the chapter is on Sylow Theorems, then it is also possible that the question was implicitly expecting finite groups. – Thomas Andrews Oct 14 '14 at 05:33
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    (Very) related: http://math.stackexchange.com/questions/768561/sort-of-simple-non-hopfian-groups This question asks if condition ($\dagger$) implies simplicity. The answer shows that it does not in general, but that it does imply simplicity for finitely presented groups. – user1729 Oct 14 '14 at 08:41
  • The point of this question was to figure out exactly what $(\dagger)$ actually means. So, in your comment, when you said $(\dagger)$, did you mean my $()$ above? You must have, because it is immediate that my $(*)$ implies simplicity. Thank you for the related post. – Doug Oct 16 '14 at 07:08

2 Answers2

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The first statement is ambiguous. One way to read "$G' \cong G$" is that it asks for an isomorphism as abstract groups, and this gives something which is not obviously equivalent to the second statement; as you say, there could be some strange group $G$ with the property that all of its quotients are trivial or abstractly isomorphic to $G$ even if some of those quotients are nontrivial. I don't know an example off the top of my head though.

Another way, which is equivalent to the second statement, is to require that the isomorphism $G' \cong G$ is not only an isomorphism of groups but an isomorphism of quotient groups; here, if $\varphi_1 : G \to G_1$ and $\varphi_2 : G \to G_2$ are two quotient groups of $G$ (the quotient maps $\varphi_1, \varphi_2$ are part of the data), then an isomorphism of quotient groups between $(G_1, \varphi_1)$ and $(G_2, \varphi_2)$ is an isomorphism $f : G_1 \to G_2$ which in addition satisfies $f \circ \varphi_1 = \varphi_2$.

Qiaochu Yuan
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The selected answer to the post linked by user1729 gives an example of a non-simple group satisfying $(*)$, and so this answers my question, with the conclusion that $(*)$ does not necessarily imply $(**)$ (of course, $(**)$ implies $(*)$!).

Doug
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