Question about $\mathbb{Z}$

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.

• 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 $m|n$. Jan 17, 2021 at 8:03
• 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. Jan 17, 2021 at 8:27
• @ArturoMagidin, can you avoid an argument by contradiction to prove that $\Bbb Z$ hasn't got finite nontrivial subgroups?
– user870827
Jan 17, 2021 at 13:17
• 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. Jan 17, 2021 at 14:58
• @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. Jan 17, 2021 at 18:21

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.