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--K. Brumbaugh, 22:50, 3 February 2009 | --K. Brumbaugh, 22:50, 3 February 2009 | ||
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| + | Show that <math>\scriptstyle U(20)\not=\langle k\rangle</math> for any <math>\scriptstyle k</math> in <math>\scriptstyle U(20)</math>. [Hence, U(20) is not cyclic.] | ||
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| + | <math>\scriptstyle U(20)={1,3,7,9,11,13,17,19}</math>. | ||
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| + | <math>\scriptstyle1^1=1\ \Rightarrow\ \langle1\rangle=\{1\}</math>. | ||
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| + | <math>\scriptstyle3^1=3,\ 3^2=9,\ 3^3=27\equiv7,\ 3^4=21\equiv1\ \Rightarrow\ \langle3\rangle=\{1,3,7,9\}</math>. | ||
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| + | <math>\scriptstyle7^1=7,\ 7^2=49\equiv9,\ 7^3=63\equiv3,\ 7^4=21\equiv1\ \Rightarrow\ \langle7\rangle=\{1,3,7,9\}</math>. | ||
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| + | <math>\scriptstyle9^1=9,\ 9^2=81\equiv1\ \Rightarrow\ \langle9\rangle=\{1,9\}</math>. | ||
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| + | <math>\scriptstyle11^1=11,\ 11^2=121\equiv1\ \Rightarrow\ \langle11\rangle=\{1,11\}</math>. | ||
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| + | <math>\scriptstyle13^1=13,\ 13^2=169\equiv9,\ 13^3=117\equiv17,\ 13^4=221\equiv1\ \Rightarrow\ \langle13\rangle=\{1,9,13,17\}</math>. | ||
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| + | <math>\scriptstyle17^1=17,\ 17^2=289\equiv9,\ 17^3=153\equiv13,\ 17^4=221\equiv1\ \Rightarrow\ \langle17\rangle=\{1,9,13,17\}</math>. | ||
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| + | <math>\scriptstyle19^1=19,\ 19^2=361\equiv1\ \Rightarrow\ \langle19\rangle=\{1,19\}</math>. | ||
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| + | From this, it is clear that none of the sets <math>\scriptstyle\langle1\rangle,\ \langle3\rangle,\ \langle7\rangle,\ \langle9\rangle,\ \langle11\rangle,\ \langle13\rangle,\ \langle17\rangle,\ \langle19\rangle</math> equal <math>\scriptstyle U(20)</math>. | ||
| + | :--[[User:Narupley|Nick Rupley]] 05:28, 4 February 2009 (UTC) | ||
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| + | --- | ||
| + | This helped out a lot. So just to be clear though, since none of the sets have an order of 8 then it's not cyclic? For it to be cyclic one of the sets must have an order of 8? | ||
| + | --[[User:Lchinn|Lchinn]] 23:52, 4 February 2009 (UTC) | ||
Latest revision as of 18:52, 4 February 2009
Does anyone know how to solve this?! Or even how to explain it? I'm having a hard time understanding this topic and was wondering if someone could explain it. Thanks! --Awika 20:09, 3 February 2009 (UTC)
What is the question? I have the 6th edition and do not know what this problem states.
--K. Brumbaugh, 22:50, 3 February 2009
Show that $ \scriptstyle U(20)\not=\langle k\rangle $ for any $ \scriptstyle k $ in $ \scriptstyle U(20) $. [Hence, U(20) is not cyclic.]
$ \scriptstyle U(20)={1,3,7,9,11,13,17,19} $.
$ \scriptstyle1^1=1\ \Rightarrow\ \langle1\rangle=\{1\} $.
$ \scriptstyle3^1=3,\ 3^2=9,\ 3^3=27\equiv7,\ 3^4=21\equiv1\ \Rightarrow\ \langle3\rangle=\{1,3,7,9\} $.
$ \scriptstyle7^1=7,\ 7^2=49\equiv9,\ 7^3=63\equiv3,\ 7^4=21\equiv1\ \Rightarrow\ \langle7\rangle=\{1,3,7,9\} $.
$ \scriptstyle9^1=9,\ 9^2=81\equiv1\ \Rightarrow\ \langle9\rangle=\{1,9\} $.
$ \scriptstyle11^1=11,\ 11^2=121\equiv1\ \Rightarrow\ \langle11\rangle=\{1,11\} $.
$ \scriptstyle13^1=13,\ 13^2=169\equiv9,\ 13^3=117\equiv17,\ 13^4=221\equiv1\ \Rightarrow\ \langle13\rangle=\{1,9,13,17\} $.
$ \scriptstyle17^1=17,\ 17^2=289\equiv9,\ 17^3=153\equiv13,\ 17^4=221\equiv1\ \Rightarrow\ \langle17\rangle=\{1,9,13,17\} $.
$ \scriptstyle19^1=19,\ 19^2=361\equiv1\ \Rightarrow\ \langle19\rangle=\{1,19\} $.
From this, it is clear that none of the sets $ \scriptstyle\langle1\rangle,\ \langle3\rangle,\ \langle7\rangle,\ \langle9\rangle,\ \langle11\rangle,\ \langle13\rangle,\ \langle17\rangle,\ \langle19\rangle $ equal $ \scriptstyle U(20) $.
- --Nick Rupley 05:28, 4 February 2009 (UTC)
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This helped out a lot. So just to be clear though, since none of the sets have an order of 8 then it's not cyclic? For it to be cyclic one of the sets must have an order of 8?
--Lchinn 23:52, 4 February 2009 (UTC)
