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Two abstract groups $G$ and $H$ are called equivalent, $G\sim H$, if each of them is isomorphic to a subgroup of another.

Question: Can a simple group $G$ be equivalent to a non-simple group $H$?

Of course, we are talking about infinite groups here. Thanks.

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    $\begingroup$ Yes it's easy, the group $A$ of finitely supported even permutations of the integers and $A\times F$ for any nontrivial finite group. $\endgroup$
    – YCor
    Commented May 13, 2017 at 12:25
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    $\begingroup$ (I didn't post as an answer because I rather answer in comments off-topic questions) $\endgroup$
    – YCor
    Commented May 13, 2017 at 12:26
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    $\begingroup$ @YCor I think you are being too strict here in considering it off-topic. The vast majority of questions on math.stackexchange are much more elementary than this, and so that site is becoming less and less suitable for questions at this level (although it is true that someone would probably have answered it there). $\endgroup$
    – Derek Holt
    Commented May 13, 2017 at 12:37
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    $\begingroup$ By definition, "MathOverflow is a question and answer site for professional mathematicians". I am one, so I feel free to ask questions to my colleagues if I feel it would take me unreasonably long time to find the answer by myself. Note that there is nothing in the definition regarding "asking vs answering" statistics and other religious beliefs. Half of the questions I asked on SE remained unanswered, and many of them are not even viewed sufficiently many times. So I need repost them on MO after losing a few days in vain. $\endgroup$
    – Bedovlat
    Commented May 13, 2017 at 13:00
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    $\begingroup$ In practice, MO has never been for technically for professional mathematicians. Questions from non-professional mathematicians are welcome when they have are of research level, while too elementary, unclear, or too broad questions of professionals are often closed or migrated too. And many users act anonymously and are not classified so. Thus, I think this sentence is a rough approximation of the target, also in order to discourage people with no relation to research to post highly off-topic questions here, but is certainly not here to select qualified users among others. $\endgroup$
    – YCor
    Commented May 13, 2017 at 13:25

1 Answer 1

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Yes. (This is corrected and expanded since the first version.) There are easy examples of simple groups $G$ such that $G\times G$ is isomorphic to a subgroup of $G$. One example is the group of finitely supported even permutations of a countable infinite set. Another is the quotient of all permutations of a countably infinite set by the finitely supported ones. Another is the infinite special linear group (direct limit of $SL_n(k)$) of a field.

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