# Infinite subgroup Product

Define the operation $$(a,b)\circ (c,d) =(ac,ad+b).$$

a) Prove that $$\left( \mathbb{Q}\setminus\{ 0\}\times\mathbb{Q} ,\circ \right)$$ is a group.

b) Let $$H$$ be an infinite subgroup of $$\left( \mathbb{Q}\setminus\{ 0\}\times\mathbb{Q} ,\circ \right)$$ that is cyclic and doesn't contain any element of the form $$(1,q) ,$$ where $$q$$ is a nonzero rational. Show that there exist two rational numbers $$a,b$$ such that $$H=\left\{ \left.\left( a^n, b\cdot\frac{1-a^n}{1-a} \right)\right| n\in\mathbb{Z} \right\}.$$

## What I thought:

1/ For the first question, it's clear that $$\circ$$ is a law of internal composition.

• associativity. Let $$(a_1,b_1),(a_2,b_2)$$ and $$(a_3,b_3)$$ three elements in $$\mathbb{Q}-\{0\}\times \mathbb{Q}$$, $$(a_1,b_1)\circ [(a_2,b_2)\circ (a_3,b_3)]=(a_1,b_1)\circ(a_2a_3,a_2b_3+b_2)=(a_1a_2a_3,a_1a_2b_3+a_1b_2+b_1)$$$$[(a_1,b_1)\circ(a_2,b_2)]\circ (a_3,b_3)=(a_1a_2,a_1b_2+b_1)\circ (a_3,b_3)=(a_1a_2a_3,a_1a_2b_3+a_1b_2+b_1)$$Hence, $$\circ$$ is associative.

• Neutral element.

• Your solution is a good start. Can you guess the neutral element? Then calculate inverse of an element $(a,b)$. Then go on to part b. – Berci Nov 29 '19 at 12:15
• Hint on the neutral element: If $(a,b)$ is the neutral element, then what does the equation for the operation look like? The first element of the pair lets you determine $a$, the second then gives you $b$. You then have to verify that what you get really is a neutral element, not only a left-neutral. – celtschk Nov 29 '19 at 12:28
• Please ask one question at a time. – Shaun Nov 29 '19 at 14:08
• @Shaun Sorry, it's two people using this account. – Benemon Nov 29 '19 at 14:13

You already checked associativity, which is good. The identity element is $$(1,0)$$, and the inverse of an element $$(a,b)$$ with $$a\neq0$$ is $$\left(\frac1a,-\frac ba\right)$$. Finally, closure follows from the fact that $$ab\neq0$$ in $$\mathbb{Q} \Leftrightarrow a,b\neq0$$, and that rationals are closed under multiplication and addition.
For b), we are given that $$H$$ is cyclic, so it is generated by a single element, say $$(a,b)$$. Then, you can prove, using induction, that repeated multiplication of the generator (or its inverse) yields $$(a,b)^n=\left(a^n,b\sum_{i=0}^{n-1}a^i\right)$$ Where $$n\in\mathbb{Z}$$. The sum on the right is just a geometric series, and it is well-known that
$$\sum_{i=0}^{n-1}a^i=\frac{1-a^n}{1-a}$$ Hence, $$H=\left\{ \left( a^n,b\frac{1-a^n}{1-a}\right) :a,b\in G, n\in\mathbb{Z}\right\}$$ Where $$G=(\mathbb{Q}\setminus\{0\}\times\mathbb{Q},\circ)$$, and $$(a,b)$$ a generator of $$H$$.