Five true or false questions on abstract algebra 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?
 A: (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.
A: 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 $|H| = |\langle a \rangle | = |a| = 51$. 
