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301.5I Cayley's Theorem for Finite Groups 

Matthew Salomone
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Symmetric groups are not only interesting to algebra; they "are" algebra, at least for finite groups: Every finite group can be realized as a subgroup of a symmetric group. Here, we suggest the proof using Cayley tables, and further show how this implies that any finite group can be represented by matrix multiplication (the left regular representation).

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14 окт 2018

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Комментарии : 15   
@viveknsharma
@viveknsharma 5 лет назад
This Video is a Masterpiece of Clarity...
@AirborneLRRP
@AirborneLRRP 5 лет назад
Mind blowing - finite groups are subgroups of Sn and can be represented as matrices. What an existence
@MatthewSalomone
@MatthewSalomone 5 лет назад
AirborneLRRP And that’s only the tip of the representation theory iceberg...
@ClumpypooCP
@ClumpypooCP 9 месяцев назад
@@MatthewSalomonemath is awesome!
@ck.mohapatra
@ck.mohapatra 4 года назад
Thanks a lot for these videos :) The amount of effort you have put into it is clearly visible. A small note: The two extreme right matrices should be swapped at instance 8:14.
@MatthewSalomone
@MatthewSalomone 4 года назад
Chandra Kanta Mohapatra Thanks, and thank you for the errata!
@rasraster
@rasraster 3 года назад
Actually, I think he had them right in the first place. The third matrix transforms [1234]T to [3421]T, which is the result of cycle (1423).
@mathdrmath9988
@mathdrmath9988 10 месяцев назад
Just excellent explanation
@normmacdonaldfan
@normmacdonaldfan 2 года назад
Mind blowing
@binaykumbhar175
@binaykumbhar175 5 лет назад
Nobody has explained group theory in RU-vid better than u sir..... Thx.... By the way it's 1=( ) -1=(1 2)(3 4) i=(1 3 2 4) -i=(1 4 2 3).
@rasraster
@rasraster 3 года назад
He actually has i and -i correct, if you check the Cayley table.
@ojas3464
@ojas3464 21 день назад
Since every Group is nonempty, Every nonempty set can be first viewed as a copy of a subset of all bijections onto itself. Since ∀ n ≥ 1 n ≤ n!, We are trying to carry the Second Inequality over to Groups with these Cardinalities. ( G ≤ Perm (G) ) Since n always divides n! we still have to resist the temptation for a converse of Fermat Theorem, since n may not be a prime. Our Group of n elements can be buried deep anywhere within a huge Group of Size n!, requiring some finesse
@joefuentes2977
@joefuentes2977 26 дней назад
Thank you guy who sounds like Ben Shapiro but who looks like Michael Carbonaro.
@ranael-achkar9936
@ranael-achkar9936 Год назад
YOU ARE THE KING OF ABSTRACT ALGEBRA!
@Mathin3D
@Mathin3D 7 месяцев назад
Nope!
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