Any two non-zero subgroups of Z are isomorphic.
Solution:
Yes, it is correct. Let "H \\leq \\mathbb{Z}" be a subgroup. If "H=\\{0\\}" then "H=0 \\mathbb{Z}" and we are done. Otherwise H must contain non-zero elements, some of them positive. Let "n=\\min \\{k>0: k \\in H\\}" . Then it can be shown that "H=n \\mathbb{Z}" . We have "n \\in H" and hence obviously "n \\mathbb{Z} \\leq H" , because H is closed to addition and additive inverses. For the other direction let "k \\in H" . We can divide k by n with remainder. We get an expression of the form k=q n+r where "q, r \\in \\mathbb{Z}, 0 \\leq r<n" . But then note that "r=k+(-q) n \\in H" . Since n is by definition the smallest positive number in H and "0 \\leq r<n" we have to conclude that r=0, and hence "k \\in n \\mathbb{Z}" . This shows that "H \\leq n \\mathbb{Z}" .
If "n \\neq 0\\ then\\ n \\mathbb{Z}" is indeed isomorphic to "\\mathbb{Z}" . We can define an isomorphism "\\varphi: \\mathbb{Z} \\rightarrow n \\mathbb{Z}\\ by\\ k \\rightarrow n k"
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