Prove set G= {1, 2, 3, 4, 5, 6} is abelian group of order 6, multiplication modulo 7 | Jayesh

[@ethicsoftechnology5220]

Love u sir . My doubts is clear . This q asked in gtu exam summer 21 q.3 C . 7 marks

[@brmr1884]

Thank you so very much sir this video helped a lot ☺️ I was literally confused with this table but now 💪🤟 .. I will definitely going to solve these problems

[@gayatrithakur4966]

Superb unbelievable you explain so easily wow thanks

[@kamalpreet_kaurr]

Very well and easy way ..each and every step is clearly explained....thank you so much..😊❤

[@samibhadgaokar2326]

Arigatto!

[@tmrxparth]

Sir I have mind but in that type of questions I was failed that's why this video is very helpful for me

[@UniversalGuru_]

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Thank you so much wholeheartedly!!!

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[@dcs-56aswanth79]

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[@gowthamikgowthamik2216]

Thank u very much sir i understand easily

[@EasyExamNotes]

Glad to hear that

[@auc_xd]

do we have to use all 5 properties to prove the set G is an abelian group?

[@ZaraQureshi-i1d]

yes that is must

[@nature_addicts21]

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[@AdithyaMenon-j7s]

Thankyousir

[@sahanab874]

Thank you sir. I understand easily

[@nitinmourya418]

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[@1CProgrammer]

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[@shreypatel6803]

THANK YOU ... PADHNE PE KUCH SAMJHA NAI BUT AAPNE SAMJHA DIYA

[@DOPEFML]

It was very helpful

[@jamkhan8853]

I understand very well

[@estationary8751]

Great...

[@AyushSharma-vh2qp]

Sir plz help me in this type of question . Show that D24 = {1, 2, 3, 4, 6, 8, 12, 24} with +, -;and/defined on D24 a + b = I em (a, b), a . b = gcd (a, b) and a' =24/a for any a,b E D24• Prove that D24 is a Boolean algebra.

[@EasyExamNotes]

Sure

[@vermastudiesmathematics3274]

Nice teaching

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[@mysteryl1775]

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[@sharonsitaram8676]

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[@abhinav6656]

Thankuh so much❤❤

[@12ekta]

Thanks, this was really helpful

[@mjmaruf275]

Thanks a lot Sir.😍

[@CS_b2arya]

Excellent💯

[@darla6706]

Tq so much sir It is very clear.... Thank You.

[@yatenderthakur3104]

a ki power 5 =e. And (ab)b inverse 1. = b square . Find order of b. Plss solve thiss ques plss sir

[@ShivaniSingh-il8ig]

gajab kya bataya thangs sir

[@MIND_.HACKER]

Very nice ser❤

[@y.ezhilarasishree2017]

Thank u

[@raiza.chakufora5271]

Do we do the same for modulo 17

[@prabhakarshelke584]

Thanks sir

[@romenmoirangthem4414]

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[@yash391]

very nice thank you sir

[@ajstudy7672]

thanku sir today is my exam

[@happypandab7136]

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[@nemaram7706]

Thanks a lot ☺🎅

[@chandubox1857]

Nice explanation

[@EasyExamNotes]

Thanks and welcome

[@Rich_song12]

I like this video sir thankyou for this video

[@rupalirupalidevi720]

Thanku sir You live long and healthy 😊😊

[@ghaidaa.s.m3922]

Thanks

[@ritika2583]

thank you

[@mustangboy3533]

Thaks ❤❤

[@shaik.......mini....3187]

Nice😊

[@fathima8118]

Thank you sir

[@EasyExamNotes]

Welcome

[@jayasribavandla1639]

Well said sir🥹🥹

[@durgaprasad3316]

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[@Ur_s_mahii]

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[@anushamodhala5215]

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[@kkalyan1442]

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[@shikhashrivastava6083]

Pls make a video on preparation of comedk

[@sabaateeq8545]

thnx

[@devavathhanmanthu479]

Nice sir

[@Ayesha-g4y]

If there is a ,set whose elements are 2,4,6,8,10 and if we make a table of it exactly like this So its closure property and associative property will be satisfied but if we look it from the perspective of identity, then it must be one, but one is not inside this set, so what?

[@shubhvichar9410]

Thq

[@romansavelyev1147]

Not sure if you guys are sarcastic😂 I did not understand a single word. But I got the idea after he drew the table.

[@anjaneyasharma322]

From 3 × 3 groups are non abelian as products are more than 6

[@nikkikumari3268]

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[@dcs-56aswanth79]

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[@memesguru1288]

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[@samreenunissa3294]

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[@aanchalkumarisingh1372]

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[@divyanshpandey676]

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[@stonecoldcold2941]

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[@jyotidevi4365]

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[@godvstruth2488]

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[@manikumar9281]

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