- #1

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Consider the map [itex] f: 2\Bbb{Z} \to 4\Bbb{Z}[/itex] defined by f(n)=2n. It's obvious that f is surjective and it's easy to verify that f is injective because if f(m)=f(n) then 2m=2n and we obtain: m=n. This map is a homomorphism of groups because:

f(m+n)=2(m+n)=2m+2n=f(m)+f(n)

so it implies that [itex](\Bbb{2Z},+)\cong (\Bbb{4Z},+)[/itex]. Is that really true? I guess it must be but it sounds a bit weird to me. This result implies that every infinite cyclic group has a subgroup of it which is isomorphic to itself!!! sounds weird and counter-intuitive to me :( Although I must admit that it's just as weird as N and Z having the same cardinal numbers so maybe it's OK? Anyway, is it true that every infinite cyclic group has a subgroup of it which is isomorphic to itself?