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Prove that every subgroup of an infinite cyclic group is characteristic.

I know that every infinite cyclic group is isomorphic to $\Bbb Z$, and any automorphism on $\Bbb Z$ is of the form $\alpha(n) = n$ or $\alpha(n) = -n$. That means that if $f$ is an isomorphism from $\Bbb Z$ to some other group $G$, the isomorphism is determined by $f(1)$. But from here I can't figure out how to show that it's characteristic.

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    $\begingroup$ Hint : Every subgroup of $\Bbb Z$ is infinite cyclic. Thus, in order to test the stability of a subgroup by the automorphisms, it suffices to test it for the generator. $\endgroup$ – paf Aug 7 '16 at 12:23
  • $\begingroup$ You sentence starting, "That means ..." is irrelevant as it talks of some other group $G$. Characteristic subgroup is a condition involving automorphisms, and not isomorphisms to other groups. So describe all subgroups and see what happens to them under the two automorphisms you ahve correctly identified. $\endgroup$ – P Vanchinathan Aug 7 '16 at 12:29
  • $\begingroup$ @paf So all subgroups are of the form $k\Bbb Z$, and by automorphisms $\alpha(k\Bbb Z) = k\Bbb Z$. Therefore, since all infinite cyclic groups are isomorphic to $\Bbb Z$ and isomorphisms map cyclic subgroups to cyclic subgroups, characteristicness is preserved? $\endgroup$ – Oliver G Aug 7 '16 at 13:01
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Every subgroup $H$ of $\mathbb{Z}$ are is in the form $H=n\mathbb{Z}$ for some $n \in \mathbb{N}$. Those are clearly stable with respect to automorphisms of $\mathbb{Z}$, which, as you said, consist of the identity automorphism and the automorphism which sends $m$ to $-m$, for every $m \in \mathbb{Z}$.

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