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I am being asked to create the subgroup lattice of the cyclic group $Z_{90}$=$\left \langle x \right \rangle$. I went ahead and found all the factors of 90 and created the diagram below. Is it done correctly? For example should $\left \langle 3 \right \rangle$ be connected with both $\left \langle 6 \right \rangle$ and $\left \langle 9 \right \rangle$? Thank you in advance for the help!

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  • $\begingroup$ You're missing a few connections, 2 also connects to 6, 3 to 15, 9 to 45, etc. $\endgroup$ – SquirtleSquad Nov 4 '16 at 23:53
  • $\begingroup$ Yes, both $\langle 6 \rangle$ and $\langle 9\rangle$ are contained in $\langle 3 \rangle$, so they are connected. In fact, all the ones you have connected should be connected, but you're missing quite a few. $5$ to $10$, for instance. $6$ to $30$ and $15$ to $30$. $\endgroup$ – Arthur Nov 4 '16 at 23:54
  • $\begingroup$ Ok, I see, thank you. I fixed it in the new image. Is it supposed to look messy with so many connections? $\endgroup$ – H.Quin Nov 5 '16 at 0:08
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The new one looks better, and, as far as I can tell, correct. It's also good to see that you've ordered it nicely in levels the way you have with $2, 3, 5$ on the first, $6, 9,10, 15$ on the second and $18, 30, 45$ on the third. Yes, it's supposed to be that many lines, but I think it would've been slightly less messy if you'd put $3$ between $2$ and $5$.

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  • $\begingroup$ Awesome, I will do that. Thank you very much for the help! $\endgroup$ – H.Quin Nov 5 '16 at 0:34
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By using unique factorization theorem, $90=2\cdot3^2\cdot5$.
I think it would be better if you draw this subgroup lattice using a cube where the 'length' is 2, 'width' is $3^2$ and 'height' is 5 so that each subgroup is connected in a clearer way as shown below:

Subgroup Lattice of $C_90$

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  • $\begingroup$ That does look very neat! Thank you! $\endgroup$ – H.Quin Nov 5 '16 at 1:05

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