$(\Bbb Z,+)$ 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 $\langle a\rangle=\langle 120\rangle ?$

  • $\begingroup$ $a$ must be the least common multiple of $24,30,12$, which is $120$. $\endgroup$ – Crostul Mar 30 '18 at 9:43
  • 1
    $\begingroup$ @Crostul 1) Why are you answering in a comment? 2) Remember that we may also have $a=-120$. $\endgroup$ – 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 $a\mathbb{Z} \cap b\mathbb{Z} \subseteq lcm(a,b)\mathbb{Z}$.

For your question you need to show these easy facts:

(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$).


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