# cyclic group

• Sep 6th 2009, 07:11 PM
MathBird
cyclic group
Need some help on the following 2 related problems:
Show that any finitely generated subgroup of the additive group of rationals (Q,+,0) is cyclic.
Prove that this group is not isomorphic to the direct product of two copies of it.
• Sep 6th 2009, 07:41 PM
ThePerfectHacker
Quote:

Originally Posted by MathBird
Need some help on the following 2 related problems:
Show that any finitely generated subgroup of the additive group of rationals (Q,+,0) is cyclic.

Remember some facts about the structure of $\displaystyle \mathbb{Z}$. For $\displaystyle x_1,...,x_l\in \mathbb{Z}$ define $\displaystyle (x_1,...,x_l)$ to be the ideal generated these elements, i.e. $\displaystyle \{\Sigma_i a_ix_i | a_i \in \mathbb{Z} \}$. Since ideals are principal it means $\displaystyle (x_1,...,x_l) = (d)$, where $\displaystyle d\geq 0$. We call $\displaystyle d$ the greatest common divisor of $\displaystyle x_1,...,x_l$.

Now given, a finitely generated subgroup $\displaystyle H$ of $\displaystyle \mathbb{Q}^+$. This means, by definition, $\displaystyle H = \left< \frac{p_1}{q_1},...,\frac{p_2}{q_2} \right>$. The subgroup generated by these elements is precisely $\displaystyle \sum_i a_i \frac{p_i}{q_i}$ where $\displaystyle a_i\in\mathbb{Z}$. Define, $\displaystyle q=\Pi_i q_i$ and $\displaystyle Q_i = \frac{q}{q_i}$. Thus, $\displaystyle \sum_i a_i\frac{p_i}{q_i} = \frac{1}{q}\sum_i a_ip_iQ_i$. Define $\displaystyle d$ to be the greatest common divisor of $\displaystyle p_1Q_1,...,p_nQ_n$. Then, as seen in the first paragraph, $\displaystyle H=\left< \frac{d}{q}\right>$. Thus, we see that $\displaystyle H$ must be cyclic.

Quote:

Prove that this group is not isomorphic to the direct product of two copies of it.
Since it is an infinite cyclic group (assuming $\displaystyle d>0$) it is isomorphic to $\displaystyle \mathbb{Z}$.
It is trivial to show that $\displaystyle \mathbb{Z}$ is not isomorphic to $\displaystyle \mathbb{Z}\times \mathbb{Z}$.
• Sep 6th 2009, 08:21 PM
MathBird
This is great!