You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
Determine whether or not U(14) is cyclic.
- Thread starter mcwang719
- Start date
Re: Cyclic???
Yes, that is the definition of cyclic. Also for future notice, state the operation in question. I know its multiplication mod 14, but only because I have the same text as you.mcwang719 said:Determine whether or not U(14) is cyclic. Justify your answer.
I'm having some trouble with this problem. So U(14)={1,3,5,9,11,13}. Do all I have to do is find a generator. If there is a generator then it's cyclic? Thanks!!!
I'm, not sure what you did there, but it looks incorrect.
If some element \(\displaystyle \alpha\) is a generator of a multiplicative group \(\displaystyle G\) with order n then \(\displaystyle G = <\alpha> = \{ \alpha^0, \,\,\alpha^1, \,\,\alpha^2, \,\, ... , \,\, \alpha^{n-1}\}\). Your job is to find one of these elements.
If some element \(\displaystyle \alpha\) is a generator of a multiplicative group \(\displaystyle G\) with order n then \(\displaystyle G = <\alpha> = \{ \alpha^0, \,\,\alpha^1, \,\,\alpha^2, \,\, ... , \,\, \alpha^{n-1}\}\). Your job is to find one of these elements.
No, did you read what I wrote? For example \(\displaystyle (U(6), \,\, \cdot (mod \,\, 6)) = <5>\) because \(\displaystyle <5>=\{5^0, \,\, 5^1, \,\, 5^2, \,\, 5^3, ... \,\, \} = \{1, \,\, 5\} = U(6)\).
What you stated is important too, but somewhat irrelevent here. The order of an element always divides the order of a (finite) group, so the order of a group will be a multiple of the order of an element. i.e. if |a|=3 then |G|=3k for some non-negative integer k. So, a^|G| = (a^3)^k = e^k = e.
What you stated is important too, but somewhat irrelevent here. The order of an element always divides the order of a (finite) group, so the order of a group will be a multiple of the order of an element. i.e. if |a|=3 then |G|=3k for some non-negative integer k. So, a^|G| = (a^3)^k = e^k = e.
mcwang719 said:Ok so I think I understand U(14) is cyclic because U(14)=<13>={13^0,13^1,13^2...}={1,13}=U(14). Is that correct? thanks
I really think you need to reread the topics on cyclic groups. U(14) is certainly not {1,13}. And <13> is defnetly not U(13). (edited)
Yes, but 13^2 = 1, 13^3 = 13, .... Thus <13>={1,13} BUT U(14) = {1,3,5,7,9,11,13}. So how does <13>=U(14)?mcwang719 said:What I did was I followed your example on U(6), so in U(14), 13^0=1, 13^1=13 and so on. Isn't that what you showed in your example with U(6)? Thanks.
My example was not to show a pattern as it seems you've picked out. So you know 13 is not a generator, keep looking.
Ok after reading the book I finally think I got it! U(14) is cyclic because it has the generators <3> and <5>.
<3>={1,3,9,13,11,5} (has all the elements)
<5>={1,5,11,13,9,3}
<9>={1,9,11} (doesn't have all the elements)
<11>={1,11,9}
<13>={1,13}
Can you guys confirm this is correct. Thanks!!!
<3>={1,3,9,13,11,5} (has all the elements)
<5>={1,5,11,13,9,3}
<9>={1,9,11} (doesn't have all the elements)
<11>={1,11,9}
<13>={1,13}
Can you guys confirm this is correct. Thanks!!!
mcwang719 said:Ok after reading the book I finally think I got it! U(14) is cyclic because it has the generators <3> and <5>.
<3>={1,3,9,13,11,5} (has all the elements)
<5>={1,5,11,13,9,3}
<9>={1,9,11} (doesn't have all the elements)
<11>={1,11,9}
<13>={1,13}
Can you guys confirm this is correct. Thanks!!!
YES! You finally got it.