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 November 19th, 2013, 06:25 AM #11 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: How to find cyclic subgroup from Cayley's Table Cyclic groups are groups that are generated by a single element. Not every abelian group is cyclic.
 November 19th, 2013, 06:29 AM #12 Newbie   Joined: Aug 2010 Posts: 27 Thanks: 0 Re: How to find cyclic subgroup from Cayley's Table means that i find subgroups generated by every single element as of e, a, a^2, a^3, b, ba, ba^2, ba^3?
November 19th, 2013, 07:15 AM   #13
Math Team

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Re: How to find cyclic subgroup from Cayley's Table

Quote:
 Originally Posted by jonbryan80 means that i find subgroups generated by every single element as of e, a, a^2, a^3, b, ba, ba^2, ba^3?
Roughly, yes. Can you find them out?

 November 19th, 2013, 05:57 PM #14 Newbie   Joined: Aug 2010 Posts: 27 Thanks: 0 Re: How to find cyclic subgroup from Cayley's Table
 November 19th, 2013, 08:19 PM #15 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: How to find cyclic subgroup from Cayley's Table Well done! Now, note that in the image below, there are examples of groups which appears twice or more. Cancel them out by listing them one by one.
 November 19th, 2013, 10:15 PM #16 Newbie   Joined: Aug 2010 Posts: 27 Thanks: 0 Re: How to find cyclic subgroup from Cayley's Table This is my solution for part b where the question is asking for all cyclic subgroups of Q4. Have I done it correctly?
 November 19th, 2013, 10:19 PM #17 Newbie   Joined: Aug 2010 Posts: 27 Thanks: 0 Re: How to find cyclic subgroup from Cayley's Table This is my solution for part c where the question is asking for cl(a) and cl(b) namely as conjugacy class of a and conjugacy class of b. Please advice. Thanks in advance.
November 19th, 2013, 10:20 PM   #18
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Re: How to find cyclic subgroup from Cayley's Table

Quote:
 Originally Posted by jonbryan80 Have I done it correctly?
Yes. The cyclic subgroups are then <a>, <a^2>, [b], <ba>, which is equivalent to what you wrote.

Quote:
 Originally Posted by jonbryan80 This is my solution for part c where the question is asking for cl(a) and cl(b)

 November 19th, 2013, 10:33 PM #19 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: How to find cyclic subgroup from Cayley's Table Now, as a last note, I would like you to inform how the question tricked you into doing hard works. The quaternions are usually defiend by the presentation : $\langle -1, i, j, k | (-1)^2= 1, i^2 = j^2 = k^2 = ijk = -1 \rangle$ But this definition is derived by setting a = x, b = y and then k = xy. I'd say you do the excersises again by using the usual definition as shown above and realize how easy it gets after that.
 November 19th, 2013, 10:50 PM #20 Newbie   Joined: Aug 2010 Posts: 27 Thanks: 0 Re: How to find cyclic subgroup from Cayley's Table thanks for your patience in helping me. may god bless u mathbalarka

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# cayley cyclic table

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