Algebra with Q/Z, finite index.

Hello teacher~

Prove that Q has no proper subgroups of finite index.

Deduce that Q/Z has no proper subgroups of finite index.

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Prove that Q has no proper subgroups of finite index.

pf)
Let N be a proper subgroup of Q of finite index.
Let [Q : N] = n. (n is a positive integer.)

Since (Q, +) is abelian group,
N is a normal subgroup.

so, |Q/N| = n.

Let g : Q -> Q, g(q) = n.q = (q+...+q)
so, g is a automorphism.
so, Q = n.Q = {n.q | q in Q}

Since |Q/N| = n,
(q + N)^n = N = q^n + N = n.q + N.
so, n.q in N for any q in Q.

so, Q = n.Q subset N subset Q.
so, Q = N.
so, contradiction.

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Deduce that Q/Z has no proper subgroups of finite index.

Sorry, I need your advice.
I know that Q/Z is a divisible group.
Of course, (Q, +) is also a divisible group.

Def)
A nontrivial abelian group G is called divisible
if for each element g in G and each nonzero integer k
there is an element x in G such that x^k = g,
(in additive notation, x.k = g)


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