Lagrange's Theorem and Index of Subgroups | Abstract Algebra

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We introduce Lagrange's theorem, showing why it is true and follows from previously proven results about cosets. We also investigate groups of prime order, seeing how Lagrange's theorem informs us about every group of prime order - in particular it tells us that any group of prime order p is cyclic and has all of its non-identity elements as generators. Thus, there is only one group (up to isomorphism) of any given prime order. We conclude with a brief discussion of the index of a subgroup H in a group G, showing that the index of a subgroup is equal to the order of the group divided by the order of the subgroup. #abstractalgebra #grouptheory

Table of Contents
0:00 Intro
1:32 Proving Lagrange's Theorem
5:06 Examples
6:42 Groups of Prime Order are Cyclic
10:07 Index of a Subgroup
11:31 Recap

Abstract Algebra Exercises:

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You are going to save so many students i swear, thank you so much!

moosophy
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Lagrange would be proud of this lesson!

punditgi
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I remember reading this proof in my lecture notes, it was SO hard to follow... you've made it so simple!

erinmeyers-tw
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Dude, this channel should be way more popular.

chimetimepaprika
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one of the best videos on lagranges theorem out there. thanks so much!

AbiBowering
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I was thinking some more about this and I don't know of any example where the identity is not 0 (for an addition style binary operator) or 1 (for a multiplication style binary operator). Theoretically you could have the identity element be 2 or 3. I'll probably write some programs to see what I can come up with.

MrCoreyTexas
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How do you find the index if given the cyclic subgroup with a permutation(123)?

raiza.chakufora
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Can you give me the link to the video about order of a group ?

aashsyed
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Anybody have an example of a counterexample to the converse of lagrange's theorem? This one will be in the back of my mind for a while. I was thinking of the Klein group {1, 3, 5, 7} with binary operator multiplication mod 8 and identity 1, I thought of expanding it to {1, 3, 5, 7, 11, 13} with binary operator multiplication mod 14, but that's not a group because if you write out a Cayley table some of the elements don't have inverses (but the binary operator is closed). You appreciate how hard it is to come up with groups trying things like this.

I think since 12 has a lot of divisors that would be the way to go looking for a counterexample, since you could look for a subgroup with 2, 3, 4, or 6 elements that doesn't exist.

MrCoreyTexas
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Hello, thank you for this helpful video!
Any advice to get full mark in this course?

Senorita-M
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wait so does it go both ways i.e. if G is a finite cyclic group then its order is a prime number and vice versa ?

itachi
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How to proof set with single element is a group ?

dr.anubhavkumarprasad