Let the group $$G= \mathbb{Z/6Z} \times \mathbb{Z/4Z}$$ I know the element $a=(1,3)$ has order $12$ and $b=(3,1)$ has order $4$. How do I determine the elements of $H:=\langle a\rangle \cap \langle b\rangle$?

  • $\begingroup$ That's only 16 elements you're looking at. (15 at most, if you don't double-count the identity.) Just enumerate them all and look for overlap? $\endgroup$ – Steven Stadnicki Jun 27 '18 at 0:34
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    $\begingroup$ Just notice $\langle a \rangle = \{(0,0),(1,3),(2,2),(3,1),(4,0),(5,3),(0,2),(1,1),(2,0),(3,3),(4,2),(5,1)\}$ and $\langle b \rangle = \{(0,0),(3,1),(0,2),(3,3)\} $. $\endgroup$ – Hugocito Jun 27 '18 at 0:35
  • $\begingroup$ Isn't there a way to do it without calculating every element? $\endgroup$ – mathie12 Jun 27 '18 at 0:39

Note that $b=a+a+a$, so $b\in\langle a\rangle$, so $\langle b\rangle\subseteq\langle a\rangle$, so $\langle a\rangle\cap \langle b\rangle=\langle b\rangle$.

  • $\begingroup$ Also, note that if this were a larger group, you might want to use the appropriate isomorphism and view $G$ either as its primary decomposition $\Bbb{Z}/(4)\times \Bbb{Z}/(2)\times\Bbb{Z}/(3)$ or its decomposition into invariant factors $\Bbb{Z}/(12)\times\Bbb{Z}/(2)$, to make the group easier to visualize. Finding the appropriate isomorphisms in this case would be a good exercise. $\endgroup$ – C Monsour Jun 27 '18 at 23:52

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