In $\mathbb{Z}_{24}$ find a generator for $\langle 21\rangle\cap \langle 10\rangle$.

My attempt:

$\langle 21\rangle = \{21, 18, 15, 12, 9, 6, 3 ,0 \}$ from adding multiples of $21\pmod{24}$ $\langle 10\rangle = \{10,20,6, 16, 2, 12, 22, 8, 18, 4 , 14, 0\}$ from adding multiples of $10\pmod{24}$

Then I take the intersection:

$ \langle 21\rangle \cap \langle 10\rangle = \{0, 6, 12, 18\}$

I think I have the intersection of subgroups right but now what do I add to each of them to check which if any are a generator? Did I do it correctly up to this point, also?

  • $\begingroup$ doesn't your second set have to include 16? $\endgroup$ – Elen Khachatryan Oct 10 '19 at 3:52
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    $\begingroup$ Thanks! Added it in $\endgroup$ – Mathstatsstudent Oct 10 '19 at 3:58

So we have:

$$\langle 21\rangle=\{21,18,15,12,9,6,3,0\}$$


$$\langle 10\rangle=\{10,20,6,16,2,12,22,8,18,4,14,0\}$$

and therefore

$$ \langle 21\rangle \cap \langle 10\rangle = \{ 18,12,6,0\}$$

Since this is a pretty small set, you can check each member if they do or do not generate it. Then, you will find that $6$ certainly does.

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  • $\begingroup$ What does it mean for 6 to generate every other element of the set? What do I add to each element to check? Is it just multiples of each respective element to check? $\endgroup$ – Mathstatsstudent Oct 10 '19 at 4:06
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    $\begingroup$ For an element to be a generator, it means that applying the operation imposed (or linked with) that set repeatedly on that element will yield you all of the members of the set. So in this case, your operation is the addition, and the repeated addition of $6$-s gives you back your set. $\endgroup$ – Elen Khachatryan Oct 10 '19 at 4:09
  • $\begingroup$ was the answer helpful? $\endgroup$ – Elen Khachatryan Oct 10 '19 at 4:14

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