Really enjoying the series so far. Just wanted to let you know that this video appears out of order in your "Group Theory" playlist. I'm guessing it was not intentional because the next video (7) introduces subgroups.
11^2 (mod15) right? since you are in U(15)... so just get 121 and subtract 15 till you get inside U(15)... you will do this 8 times or, 121 - 15 x 8 = 1 tks for watching the videos
I think it needs some correction. The order of a group is its carnality for example the order of Z10 is 10 these are {0,1,2,3,4,5,6,7,8,9} the numbers 1,3,7,9, ( the relative primes of 10) are not the orders . They are the group generators.
small mistake 8:38 16 modulo 15 should be 1 But i understand you meant to say the order of 4 is 2. Great video. So much better than reading an arcane textbook.
I am giving you a link in the video description ok? Its just the elements of Z_n "relatively prime" to n. Or the "units" of Z_n Imagine you have Z_14 = {0, 1, 2,...13} right? But now you will take only the elements relatively prime to 14, so that will be: U(14)={1,3,5, 9,13} Not the "prime numbers", e.g., check that 7 is not there. I will make a video for that... you are not the first person asking that...
U(10) is the set elements under the operation multiplication mod 10 such that each element has an inverse. We can't use the entire set Z10 { 0,1,2,3,4,.. 9} for this group because for example 2x = 1 has no solution in mod 10. But 3x =1 has a solution in mod 10, namely x = 7. Recall that for a set to be a group each element must have an inverse. It turns out, the elements of U(n) are the set of elements relatively prime to n.
Your videos are very helpful. Your explanations are clear and understanding. Thank you very much!
tks for watching my videos.
Really enjoying the series so far. Just wanted to let you know that this video appears out of order in your "Group Theory" playlist. I'm guessing it was not intentional because the next video (7) introduces subgroups.
Tks a Lot Dimitri, I will check that....
Please check if it is ok now
LadislauFernandes It is perfect now. Thank you very much.
Dimitri Tishchenko I am the one who is grateful
11^2 (mod15) right? since you are in U(15)... so just get 121 and subtract 15 till you get inside U(15)... you will do this 8 times
or, 121 - 15 x 8 = 1
tks for watching the videos
I think it needs some correction. The order of a group is its carnality for example the order of Z10 is 10 these are {0,1,2,3,4,5,6,7,8,9} the numbers 1,3,7,9, ( the relative primes of 10) are not the orders . They are the group generators.
small mistake
8:38 16 modulo 15 should be 1
But i understand you meant to say the order of 4 is 2.
Great video. So much better than reading an arcane textbook.
thank you... modern mathematics made easy...feeling relaxed..
your videos are really great.please make videos on metric spaces and topology covering proofs as well as practice problems .thanks in advance
regards
I am giving you a link in the video description ok?
Its just the elements of Z_n "relatively prime" to n.
Or the "units" of Z_n
Imagine you have Z_14 = {0, 1, 2,...13} right? But now you will take only the elements relatively prime to 14, so that will be:
U(14)={1,3,5, 9,13}
Not the "prime numbers", e.g., check that 7 is not there.
I will make a video for that... you are not the first person asking that...
yes , since 11 and 14 are relativity prime , this means that there is no number which divide both 11 and 14 except 1.
this much helping in my study
i know its a stupid question but , how can i find U(n) i m not getting it
With the order of a group, is it the number of distinct elements or is it just the number of elements?
Thank you😊
dear sir in your video you write u(10)={1,3,5,7,9} what does it called and how this set becomes kindly reply
am really enjoying the lesson thank you
So, why don't you actually explain how 7^1=7, 7^2=4, etc... wtf... Why I am so fucking stupid?
modular multiplication, so 7^2 is being represented as 7*7=49 but we're in modulo 15 so 15*3=45 making 49-45=4
Great 🙌
IB students that sutdy from here for the exams os n14 had the first question all correct
Is there any shortcut to find order
Great sir how to find number of subgroup of order 2 of a group of order12
Please check this, hope that helps:
groupprops.subwiki.org/wiki/Subgroup_structure_of_groups_of_order_12
Notes lo vunna vidham ga chepparu but explanation correct ga evvale
Sorry to say this
3:30 minus 1 thats 3 quick maths.
every man's in the fridge
i think i have to follow joseph gallian contemporary abstract algebra
I am looking for tutor.. Is there anyone who can help me for abstract algebra or other math subjects
thank sir what is the order of [24]
what do the elements of U(10) come from. why are those elements 1,3,7,9? Thanks
+YaWePcGame Mod 10 for multiplication. 3*3 = 9, 9*3 = 27 = 7 mod 10, 7*3 = 21 = 1 mod 10, 1 * 3 = 3
U(10) is the set elements under the operation multiplication mod 10 such that each element has an inverse. We can't use the entire set Z10 { 0,1,2,3,4,.. 9} for this group because for example 2x = 1 has no solution in mod 10. But 3x =1 has a solution in mod 10, namely x = 7. Recall that for a set to be a group each element must have an inverse. It turns out, the elements of U(n) are the set of elements relatively prime to n.
thanks
Questions of this ans plz?
would 11 be in U(14) ?
Yes
Thanks for your video!
Just want to add a correction to the order of 4 at 8:30
16 modulo 15 is 1. so |4| = 1 , and not 2
Jon Tan 4^2 = 16 = 1 (mod 15)
16 mod 15 is 1, but the order is 2 because 4 needs to be raised to the power of 2 (mod 15).
Tq
How do you
اشرح بالعربي افضل 😉😉
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