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If every quotient group of $G$ by a non-trivial normal subgroup is finite, then $G$ is finite.

This a statement that I'm supposed to prove if it's true or not.

If $G = (\mathbb{Z}, +)$, then all non-trivial subgroups of $G$ are $n\mathbb{Z}$ with $n = 2,3,4,\dots\,$.

But as $G$ is abelian, so every subgroup is normal and the quotient group $\mathbb{Z}/n\mathbb{Z}$ is formed by $n$ elements. So this would be a counter-example to the previous statement. Is this a valid counterexample?

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    $\begingroup$ Yes, that's a correct counterexample. $\endgroup$
    – Qiaochu Yuan
    Commented Oct 9, 2023 at 20:43
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    $\begingroup$ Note: $n=1$ is valid too. In that case $\mathbb{Z}/\mathbb{Z}$ is the trivial group with only the identity element, which is of course finite. :-) $\endgroup$
    – Sammy Black
    Commented Oct 9, 2023 at 20:46
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    $\begingroup$ From the solution-verification tag wiki: "A question with this tag should include an explanation for why the argument presented is not convincing enough." $\endgroup$
    – Shaun ♦
    Commented Oct 9, 2023 at 20:48
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    $\begingroup$ @Shaun: the take that you quote on the solution-verification tag is nonsensical. I wanted to push back on it, but it is clearly received dogma that is not worth arguing about. In this case, it is quite clear which part of the solution the OP is uncertain about. $\endgroup$
    – Rob Arthan
    Commented Oct 9, 2023 at 23:08
  • $\begingroup$ There are links to meta discussions of the solution-verification tag in its wiki, @RobArthan; I suggest you take up any complaints there. $\endgroup$
    – Shaun ♦
    Commented Oct 10, 2023 at 10:44

1 Answer 1

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Yes. $\,\,\,\,\,\,\,\,\,\,\,\,$

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