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A basic consequence of the first isomorphism theorem is that a finite group G is simple if and only if its only homomorphic images are G and the trivial group (up to isomorphism). However, I'm not sure whether or not this generalizes to infinite groups. If it doesn't, then that must mean there exists an infinite group G with at least one non-trivial proper normal subgroup, such that for each proper normal subgroup H of G, G is isomorphic to G/H.

So does this equivalent definition generalize to infinite groups?

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2 Answers 2

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Let $D$ denote the dyadic rationals- rational numbers of the form $\frac{a}{2^n}$ for some integer $n$. Then I claim that $G = D/\mathbb{Z}$ is a counterexample.

Since $G$ is abelian, every subgroup of $G$ is normal. Let $D_k \subseteq D$ be the subgroup of rational numbers of the form $\frac{a}{2^k}$. It is not hard to show that this is a complete list of subgroups of $D$ containing $\mathbb{Z}$, and it follows that all subgroups of $G$ are of the form $D_k/\mathbb{Z}$. We have $G/(D_k/\mathbb{Z}) \cong D/D_k$, and the isomorphism $D/D_k \cong D/\mathbb{Z}$ is given by the map $\phi:D \rightarrow D$ of multiplication by $2^k$.

Edit: This is the same as egreg's example for p = 2.

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  • $\begingroup$ I think that might be a counter-example, I'm not sure how to prove it though. $\endgroup$
    – Nick
    Dec 12, 2016 at 21:22
  • $\begingroup$ Made some edits $\endgroup$
    – Alex Zorn
    Dec 12, 2016 at 21:29
  • $\begingroup$ Cool. Thanks for the response $\endgroup$
    – Nick
    Dec 12, 2016 at 21:38
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If $\mathbb{Z}(p^\infty)$ denotes the Prüfer $p$-group, then for each (normal) proper subgroup $H$, $\mathbb{Z}(p^\infty)/H\cong \mathbb{Z}(p^\infty)$. Note that this group is very far from being simple.

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  • $\begingroup$ Interesting. Thanks! $\endgroup$
    – Nick
    Dec 12, 2016 at 21:28

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