I was reading about minimal and maximal subgroups and in one example it is mentioned that the additive group $\mathbb{Z} $ does not have minimal subgroups. I tried to prove this by assuming that $ \mathbb{Z} $ does have minimal subgroups to arrive at a contradiction, starting from the fact that the subgroups of $\mathbb{Z} $ are cyclic but it doesn't get anywhere.

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    $\begingroup$ It always mystifies me why the first thing people seem to try is an argument by contradiction... Note that the subgroups of $\mathbb{Z}$ are all of the form $\langle n\rangle$ for some $n\geq 0$, and that $\langle n\rangle\subseteq \langle m\rangle$ if and only if $m|n$. $\endgroup$ Jan 17, 2021 at 8:03
  • $\begingroup$ Hint: If $H$ is a non-trivial subgroup, show that $2H=\{2x\mid x\in H\}$ is a proper subset of $H$ and a non-trivial subgroup. $\endgroup$ Jan 17, 2021 at 8:27
  • $\begingroup$ @ArturoMagidin, can you avoid an argument by contradiction to prove that $\Bbb Z$ hasn't got finite nontrivial subgroups? $\endgroup$
    – user870827
    Jan 17, 2021 at 13:17
  • $\begingroup$ Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$
    – Shaun
    Jan 17, 2021 at 14:58
  • $\begingroup$ @user: You can prove it by contrapositive: if $H$ is a nontrivial subgroup, then it is infinite. In any case, I don’t object to proofs by contradiction per se. It just mystifies me that it seems to be the very first thing a lot of people try always for everything. $\endgroup$ Jan 17, 2021 at 18:21

1 Answer 1


So, as pointed out already by @ArturoMagidin, any subgroup is of the form $\langle n\rangle$. But there are always multiples of a given $n$ (in fact infinitely many: $n,2n,3n,\dots$). Now the (cyclic) subgroup $\langle kn\rangle\lt\langle n\rangle$ for any $\Bbb Z\ni k\gt1$. Thus $\langle n\rangle$ is not minimal.


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