Group Theory 6, order of a Group, order of an element

[@LadislauFernandes]

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

[@HAWI-SO]

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.

[@LadislauFernandes]

tks for watching my videos.

[@koozdra]

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.

[@LadislauFernandes]

Tks a Lot Dimitri, I will check that....

[@LadislauFernandes]

Please check if it is ok now

[@koozdra]

LadislauFernandes It is perfect now. Thank you very much.

[@LadislauFernandes]

Dimitri Tishchenko I am the one who is grateful

[@LadislauFernandes]

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...

[@xoppa09]

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.

[@liltv2012]

Your videos are very helpful. Your explanations are clear and understanding. Thank you very much!

[@fawzyhegab]

yes , since 11 and 14 are relativity prime , this means that there is no number which divide both 11 and 14 except 1.

[@Money__Matters007]

your videos are really great.please make videos on metric spaces and topology covering proofs as well as practice problems .thanks in advance regards

[@smsthea2z973]

i know its a stupid question but , how can i find U(n) i m not getting it

[@HashimAli-2013]

Great 🙌

[@cscor]

thank you... modern mathematics made easy...feeling relaxed..

[@ib4609]

With the order of a group, is it the number of distinct elements or is it just the number of elements?

[@Scripter]

3:30 minus 1 thats 3 quick maths.

[@tanishamandal6906]

every man's in the fridge

[@LokmanMahazer]

this much helping in my study

[@manikanta5551]

Is there any shortcut to find order

[@mansoorakhtar6181]

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

[@arundhatipandey1368]

Thank you😊

[@owusualfredkwabena8929]

am really enjoying the lesson thank you

[@albron2189]

Great sir how to find number of subgroup of order 2 of a group of order12

[@LadislauFernandes]

Please check this, hope that helps: groupprops.subwiki.org/wiki/Subgroup_structure_of_groups_of_order_12

[@chunkylover5367]

So, why don't you actually explain how 7^1=7, 7^2=4, etc... wtf... Why I am so fucking stupid?

[@m35926]

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

[@emprendeperez]

IB students that sutdy from here for the exams os n14 had the first question all correct

[@CallMeMantou]

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

[@yatmingcheung2772]

Jon Tan 4^2 = 16 = 1 (mod 15)

[@andrewsantamonica6025]

16 mod 15 is 1, but the order is 2 because 4 needs to be raised to the power of 2 (mod 15).

[@YaWePcGame]

what do the elements of U(10) come from. why are those elements 1,3,7,9? Thanks

[@RkeyNirun]

+YaWePcGame Mod 10 for multiplication. 3*3 = 9, 9*3 = 27 = 7 mod 10, 7*3 = 21 = 1 mod 10, 1 * 3 = 3

[@xoppa09]

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.

[@owusualfredkwabena8929]

thank sir what is the order of [24]

[@Maths_Magic_]

i think i have to follow joseph gallian contemporary abstract algebra

[@you0are0rank]

would 11 be in U(14) ?

[@Nostalgic_artist_Neha]

Yes

[@user-mz7fh2hy6x]

I am looking for tutor.. Is there anyone who can help me for abstract algebra or other math subjects

[@renupanthangi7029]

Notes lo vunna vidham ga chepparu but explanation correct ga evvale Sorry to say this

[@LokmanMahazer]

thanks

[@yakhilijamatia9377]

Questions of this ans plz?

[@manisharma421]

Tq

[@birdhanhansdah8580]

How do you

[@Tradexoffecials]

Worst teacher

[@LadislauFernandes]

I agree

[@kingomar3668]

اشرح بالعربي افضل 😉😉