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Let $\mathbb{N}_0$ denotes the set of nonnegative integers and $$ S=\left\{ \sum_{i=0}^k a_i 2^i:k \in \mathbb{N}_0,a_i \in \mathbb{Z}\,\,\forall i \right\} $$ Define $$ G=\left\{ \pmatrix {2^a &2^bx \\ 0 & 1}:a,b\in \mathbb{Z}, x \in S \right\} $$ Show that $G$ with matrix multiplication as operation is a group. I failed to show that $G$ is closed under multiplication. I don know how to show the $a_{12}$ is of the form $2^bx$. Anyone can guide me?

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I fail to see why $S\ne\mathbb{Z}$. – yohBS Mar 4 '13 at 14:36
@yohBS: can you elaborate further ? What makes you think that $S=\mathbb{Z}$? – Idonknow Mar 4 '13 at 15:17
Obviously, $S \subset \mathbb{Z}$. To see the other direction, choose $a_i = 0$ for all $i \geq 1$, and then choose $a_0 = x$ and let $x$ go through $\mathbb{Z}$. – gnometorule Mar 4 '13 at 15:55
@YACP:but that is what given in the question. I just copy it. – Idonknow Mar 5 '13 at 2:54
up vote 6 down vote accepted

$a_{12}$ can be considered as: $$2^{a+b'}x'+2^bx$$ in which $x=\sum_0^ka_i2^i$ for $k\in\mathbb N_0, a_i\in\mathbb Z,~~~~~ x'=\sum_0^k'a_j2^j$ for $k'\in\mathbb N_0, a_j\in\mathbb Z$. Let $2^{a+b'}>2^b$, so $$2^{a+b'}x'+2^bx=2^b\left(\sum_0^{k'}2^{a+b'-b+j}a_j+\sum_0^ka_i2^i\right)$$ and by $x^*=\sum_0^{k'}2^{a+b'-b+j}a_j+\sum_0^ka_i2^i$ we have $a_{12}=2^bx^*$

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$+1\quad\ddot\smile\quad$ – amWhy Mar 4 '13 at 15:03
how do we know $x^* \in S$ ? – Idonknow Mar 4 '13 at 15:06
Idonknow: First of all, we should think about what @yohBS noted you. We should think about an answer for him/her. And secondly, we need some handy calculations and rearrangement to see that $x^*$ is in $S$. – Babak S. Mar 4 '13 at 15:13
+1...and I almost got dizzy of so many upper and lower indexes! – DonAntonio Mar 4 '13 at 17:03
@DonAntonio: Thanks Don for your consideration. Honestly, I was too. :-) – Babak S. Mar 4 '13 at 17:08

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