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Proof by Bézout's identity
Let $$G = \langle g \rangle$$ and $$H$$ one of its subgroups.
 * $$h \in H \Rightarrow h \in G \Rightarrow h = g^i; i \in \mathbb{Z}$$

Consider now:
 * $$g^a, g^b \in H \Rightarrow g^{a+b} \in H$$

Bézout's identity states that there is a pair of integers $$x, y$$ that make teh following true:
 * $$ax + by = gcd(a, b) \Rightarrow g^{gcd(a, b)} \in H$$

Which gives:
 * $$H = \langle h_1, h_2, h_3, \cdots \rangle$$
 * $$H = \langle g^{i_1}, g^{i_2}, g^{i_3}, \cdots \rangle$$
 * $$H = \langle g^{gcd(i_1, i_2, i_3, \cdots)} \rangle$$

proving that all subgroups of $$G$$ are cyclic.

Now for teh divisor part, let:
 * $$|G| = |\langle g \rangle| = n \Leftrightarrow g^n = 1$$

Then:
 * $$k|n \Rightarrow (g^{\frac{n}{k}})^k = 1$$

Giving:
 * $$|\langle g^{\frac{n}{k}} \rangle| = k$$

=op