# Find a generator of the intersection of three subgroups of the integers

$(\Bbb Z,+)$ is infinite group and cyclic with $a\in(\Bbb Z,+)$.

$\langle a \rangle=\langle 24 \rangle\cap \langle 30 \rangle \cap \langle 12 \rangle$.

I want to find all of $a$ and I know $$\langle 24\rangle =\{\dots ,-96,-72,-48,-24,0,24,48,72,96,\dots \},$$

$$\langle 30 \rangle =\{\dots,-120,-90,-60,-30,0,30,60,90,120,\dots\},$$ and $$\langle 12\rangle=\{\dots ,-60,-48,-36,-24,-12,0,12,24,36,48,60,\dots\}$$

then $\langle a\rangle=\langle 120\rangle ?$

• $a$ must be the least common multiple of $24,30,12$, which is $120$. – Crostul Mar 30 '18 at 9:43
• @Crostul 1) Why are you answering in a comment? 2) Remember that we may also have $a=-120$. – Arthur Mar 30 '18 at 9:46

@Crostul already wrote but maybe it can be good to show general form. Let $a,b \in \mathbb{Z}$ so we want to show $\langle a\rangle \cap \langle b\rangle = a\mathbb{Z} \cap b\mathbb{Z} = lcm(a,b)\mathbb{Z}.$ Obviously $lcm(a,b) \in a\mathbb{Z} \cap b\mathbb{Z}$ so $lcm(a,b)\mathbb{Z} \subseteq a\mathbb{Z} \cap b\mathbb{Z}.$ To other way, by the definition of $lcm(a,b)$every multiple of a and b must be multiple of $lcm(a,b)$.So $a\mathbb{Z} \cap b\mathbb{Z} \subseteq lcm(a,b)\mathbb{Z}$.
(1)$(\forall n \in \mathbb{N})$ $a_{1}\mathbb{Z} \cap ...\cap a_{n}\mathbb{Z} = lcm(a_{1},...,a_{n})\mathbb{Z}.$
(2)$n\mathbb{Z}=m\mathbb{Z}$ iff ($n=m$ or $n=-m$).