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Let $a$, $b$ be fixed positive integers and $$H=\{ax+by \mid x,y\in \Bbb Z\}.$$ Show that $H$ is a cyclic group with $\gcd(a,b)$ as a generator.

Approach:

Let $d=\gcd(a,b)$

then $d|a, ~d|b$ i.e $d\alpha =a, ~d\beta =b$ for some integers $\alpha, \beta$

Let $ax+by\in H$ then $ax+by=d(x\alpha + y\beta)\in dZ$ i.e $$H\subset d\Bbb Z ~~~~~~~~~~~~~(1)$$

We now show that $d \Bbb Z \subset H$

By Euclidean algorithm, there exist $u, v$ such that

$ua+vb=\gcd(a,b)=d$

Please correct my steps and help me to complete the remaining part of solution. I am unable to proceed further.

EDIT: Question in clear form:

I would like to show that $$d\mathbb Z =H.$$ I have shown $$H\subset d\Bbb Z$$ and I am unable to show $$d \Bbb Z \subset H$$ Please help me to show the desired part ($d \Bbb Z \subset H$).

Sebastiano
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rama_ran
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  • One useful fact is that the greatest common divisor of any two integers is a linear combination of those integers. This follows from the Euclidean algorithm by reversing the steps. – CyclotomicField Oct 20 '22 at 23:20

2 Answers2

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First show that $H$ is a subgroup of $\mathbb Z$ which is easy to show. Then use the fact (?) that any subgroup of $\mathbb Z$ is of the form $n\mathbb Z$ for some integer $n$. Hence $H$ is a cyclic group, of the form $d\mathbb Z$ for some $d\in\mathbb Z$.

We claim $d=gcd(a,b)$.

Note that $a=a.1+b.0\in H$ and $b=a.0+b.1\in H$ so $a,b\in d\mathbb Z$ implying $d|a,d|b$ implying $d$ is a common divisor of $a$ and $b$.

Now $d\in d\mathbb Z$ so there exist integers $x,y$ so that $d=ax+by$. Let $k$ be a common divisor of $a$ and $b$ then this equation shows that $k$ divides the RHS, so $k$ divides LHS i.e. $k|d$.

These conclude that $d=gcd(a,b)$.

Landon Carter
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  • Thanks for your post. I would be very happy if you kindly suggest me the required steps to show that $$d\mathbb Z =H.$$ I have shown $$H\subset d\Bbb Z$$ and I am unable to show $$d \Bbb Z \subset H$$ Please help me to show the desired. – rama_ran Sep 25 '15 at 18:13
  • @rama_ran Let $S := \left{ a u + b y | a u + b v > 0, u \in \mathbb{Z}, v \in \mathbb{Z} \right}$. Then $S$ is nonempty since $a^2 + b^2 \in S$. Now, by the well-ordering principle, one can find $d = a x + b y$ for some $x , y \in \mathbb{Z}$. To show $d = \mathrm{gcd} \left( a , b \right)$, we can write $a = q d + r$ using the division algorithm. So $r = a \left( 1 - q x \right) + b \left( - q y \right)$. If $r > 0$ then $r \in S$, which is a contradiction. So $r = 0$ and $a = q d$. Similiary, $d | b$. If $c$ is a common positive divisior of $a$ and $b$, $c \le d$. Thus $d$ is the gcd. – Kijeong Lim Mar 02 '24 at 11:27
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Let $d=\gcd(a,b)$. Then $a=d\cdot r$ and $b= d\cdot s$ for some integers $r, s$.

Now $\gcd(r,s)=1$. Then $\exists\ r\cdot u+s\cdot v=1$ for some integers $u, v$.

So for any integer $z,\ z=r\cdot u\cdot z+s\cdot v\cdot z$.

Hence $d\cdot z= d\cdot r\cdot u\cdot z+d\cdot s\cdot v\cdot z=a\cdot u\cdot z+b\cdot v\cdot z=a\cdot x+b\cdot y$, where $x=u\cdot z$ and $y=v\cdot z$ are two integers. Hence $d\Bbb Z \subset H$.

darij grinberg
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Math-Boy
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