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Which of the following statements are true:
1. every countable group G has only countably many distinct subgroups.
2. any automorphism of the group $\mathbb{Q}$ under addition is of the form $x→qx$ for some $q\in\mathbb{Q}$
3. all non-trivial proper subgroups of $(\mathbb{R},+)$ are cyclic.
4. every infinite abelian group has at least one element of infinite order.
5. there is an element of order $51$ in the multiplicative group $(\mathbb{Z}/103\mathbb{Z})^*$

My thoughts:
1. true as union of uncountable number of countable set is uncountable
2. true as any homomorphism must be one of those form
3. false as $(\mathbb{Q},+)$ is not cyclic.
4. false example circle group.
5. true by Fermat's little theorem.

Are my guesses correct?

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5 is true, but I don't think Fermat is involved. 4 is false, but I don't know what the circle group is --- if its the group of rotations of the circle, it certainly has elements of infinite order. – Gerry Myerson Dec 10 '12 at 3:56
Your justification of (1) is weird, to say the least: what has that to do with the question asked?. (2) This is no mathematical justification at all. (3) Good. (4) Not good, as the circle group has lots of elements of infinite order. You can take the Prufer group, though. (5) Well, yes...but you can't apply Cauchy's Theorem directly as $\,51\,$ is not a prime – DonAntonio Dec 10 '12 at 4:22

(1) Let $G$ be the direct sum of copies of $\Bbb Z/2\Bbb Z$ indexed by $\Bbb N$. $G$ is countably infinite, but each subset $A\subseteq\Bbb N$ generates a distinct subgroup $G_A=\{x\in G:x_n=0\text{ for all }n\in\Bbb N\setminus A\}$ of $G$, and $\Bbb N$ has uncountably many subsets.

(4) The statement is indeed false; the group $G$ above is an infinite Abelian group in which every non-zero element has order $2$. However, the circle group is not a counterexample: $e^{i\theta}$ has infinite order iff $\frac{\theta}{2\pi}$ is irrational, so the circle group has uncountably many elements of infinite order and only countably many of finite order.

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does Q/Z provide a counterexample for (4)? – jim Dec 10 '12 at 10:58
@jim: Yes: $n\left(\frac{m}n+\Bbb Z\right)=0+\Bbb Z$. – Brian M. Scott Dec 10 '12 at 11:00

For part 5, note that $$|(\mathbb{Z}/103\mathbb{Z})^*| = \phi(103) = 102$$ and since $(\mathbb{Z}/103\mathbb{Z})^*$ is an abelian group we know that there exists a subgroup for every divisor of $102$. Hence we know that there exists some subgroup, $H$, such that $|H| = 51$. Now if you know that $(\mathbb{Z}/103\mathbb{Z})^*$ is actually cyclic, and since each subgroup of a cyclic group is also cyclic, there exists some $a \in (\mathbb{Z}/103\mathbb{Z})^*$ such that $\langle a \rangle = H$. Thus $|H| = |\langle a \rangle | = |a| = 51$.

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protected by user26857 Dec 2 '15 at 22:35

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