Is $aZ \cup b Z = \gcd(a, b)Z$?

Question

Is $aZ \cup b Z = \gcd(a, b)Z$?

I know that $\gcd(a,b)$ is a factor of both $a$ and $b$, but I am not sure how this translates to $aZ \cup b Z = \gcd(a, b)Z$. Therefore I tried to find a counter example to disprove this, but was unsure how to structure the sets $a$ and $b$, then determine a $\gcd$ from that.

Second question,

Can you prove that $aZ \cap bZ = lcm(a,b)Z$?

I know that $aZ$ and $bZ \in lcm(a,b)Z$, but I don't know how to prove this equality is equal.

Answer 1

As a rule $a\mathbb{Z}\cup b\mathbb{Z}$ is not a group. For example, $$ 2\mathbb{Z}\cup 3\mathbb{Z}=\{0,\pm2,\pm3,\pm4,\pm6,\pm8,\ldots\} $$ This is not closed under addition since $2+3=5\notin 2\mathbb{Z}\cup 3\mathbb{Z}$.

On the other hand, $a\mathbb{Z}+b\mathbb{Z}=\gcd(a,b)\mathbb{Z}$. Indeed, let $d$ be the minimal positive element in $a\mathbb{Z}+b\mathbb{Z}$. Then, $d\mathbb{Z}$ is a subgroup of $a\mathbb{Z}+b\mathbb{Z}$. On the other hand, if $k\in a\mathbb{Z}+b\mathbb{Z}$, then use the division algorithm to write $k=dq+r$ for $0\leq r<d$. Then, $r=dq-k\in a\mathbb{Z}+b\mathbb{Z}$ forcing $r=0$ by the minimality of $d>0$.

Now we have that $a\mathbb{Z}+b\mathbb{Z}=d\mathbb{Z}$, so we need to check that $d=\gcd(a,b)$. Well, as $a,b\in d\mathbb{Z}$ we have that $d|a$ and $d|b$. Now, if $c|a$ and $c|b$, we need to show that $c|d$. But, $d=ax+by$ for some $x,y\in \mathbb{Z}$ since $d\in a\mathbb{Z}+b\mathbb{Z}$. As $c$ divides both $ax$ and $by$, $c$ divides $ax+by=d$.

Answer 2

if $gcd(a, b)$ $\ne$ $a$ and $gcd(a,b)$ $\ne$ b. Then $gcd(a,b) < a$ and $gcd(a,b) < b$ so $gcd(a,b)$ is not a multiple of $a$, so $gcd(a,b)$ $\notin$ $aZ$. Likewise $gcd(a,b)$ $\notin$ $bZ$, so $gcd(a,b)$ $\notin$ $zA$ $\cup$ $bZ$. But $gcd(a,b)$ $\in$ $gcd(a,b)Z$. So $gcd(a,b)Z$ is not a subset of $aZ$ $\cup$$ bZ$ and $aZ$ $\cup$ $bZ$ $\ne$ $gcd(a,b)Z$.

(However $aZ $$\cup$ $bZ$ $\subset$ $gcd(a,b)Z$ by similar argument.)

Your second question: if$ m $$\in$ $aZ$ $\cap$$ bZ$, then $m$ is a multiple of $a$ and $m$ is a multiple of $b$ so $m$ is a multiple of $lcm(a,b)$ so $\in$ $lcm(a,b)Z$. and $\in$ $aZ$ $\cap$ $bZ$ $\subset$ $lcm(a,b)$. If $n$ $\in$ $lcm(a,b)$ then $n$ is a multiple of $lcm(a,b)$ so it is a multiple of both $a $and of $b$ so n $\in$ aZ $\cap$ bZ. So $lcm(a,b)Z$ $\subset$ $aZ$ $\cap$$ bZ$. So, yes, $aZ$ $\cap$ $bZ = lcm(a,b)Z$.