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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.

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  • $\begingroup$ 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. $\endgroup$ – Berci Nov 29 '19 at 12:15
  • $\begingroup$ 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. $\endgroup$ – celtschk Nov 29 '19 at 12:28
  • $\begingroup$ Please ask one question at a time. $\endgroup$ – Shaun Nov 29 '19 at 14:08
  • $\begingroup$ @Shaun Sorry, it's two people using this account. $\endgroup$ – Benemon Nov 29 '19 at 14:13
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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 $H$.

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