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301.3A Finite Groups and Order(s) 

Matthew Salomone
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Order of a group (the cardinal number of its elements) and order of an element in a group (the smallest nontrivial power which equals the identity).

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30 сен 2024

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Комментарии : 10   
@lotusbackground7220
@lotusbackground7220 5 лет назад
You are great professor. You make it very easy to understand by explaining visually and metaphorically. Thank you so much for taking your time to do these things!
@MatthewSalomone
@MatthewSalomone 5 лет назад
lotus background Thanks, it's my pleasure!
@wickz7119
@wickz7119 2 года назад
I’m taking abstract algebra in the fall, looking to get ahead and these lectures are SO helpful. Thank you!
@PunmasterSTP
@PunmasterSTP 2 месяца назад
How'd the class go?
@islamandmuslim6658
@islamandmuslim6658 4 года назад
RU-vid channel i would love to prescribe my friends. You are The Professor.
@PunmasterSTP
@PunmasterSTP 2 месяца назад
Did you ever prescribe this channel to some of your friends?
@PunmasterSTP
@PunmasterSTP 2 месяца назад
Alright, can't wait till we get to The Monster!
@maxpercer7119
@maxpercer7119 2 года назад
subgroups are to groups as subspaces are to vector spaces
@bonbonpony
@bonbonpony 3 года назад
I love the shoes & socks metaphor! :D Hmm… but I think that there is one more property that should me mentioned, that I think can be deduced from the group axioms: inverses are always commutative (a·a⁻¹ = a⁻¹·a = 1), so it doesn't matter if we cancel them on the right or on the left. Suppose that we wanted to cancel `a` from both sides of this: a·b = a·c but we had only the RIGHT inverse for `a`. It may seem at first that we are out of luck, because we have `a` on the left in our equation, and nothing to multiply it on its left to cancel it. But using the socks & shoes trick, we can inverse both sides of this equation, which in turn should push our common factor to the right (also inverting it as a side effect, but that's not a problem at all!): (a·b)⁻¹ = (a·c)⁻¹ b⁻¹·a⁻¹ = c⁻¹·a⁻¹ and now we have `a` on the right, so we can multiply it by its RIGHT inverse from the RIGHT :) (b⁻¹·a⁻¹)·a = (c⁻¹·a⁻¹)·a Associativity allows us to move the parentheses: b⁻¹·(a⁻¹·a) = c⁻¹·(a⁻¹·a) and cancel the inverses into identities: b⁻¹·1 = c⁻¹·1 which, as we know, leave the elements unchanged, so: b⁻¹ = c⁻¹ And inverting the inverses one more time gives us the original equation back, with all the `a`s cancelled ;) b = c So it doesn't really matter whether our inverses are LEFT inverses or RIGHT inverses: we can cancel common factors anyway :) Kinda cancelling them "on the other side of the mirror" ;) So cool! ;) But this makes me think: What do we have to do to (or which of the axioms to remove) if we wanted to actually BREAK algebra? :q Because something tells me that we should be still able to do a limited form of algebra even without the associative property.
@maxpercer7119
@maxpercer7119 2 года назад
Some authors (e.g. graduate level) define a group as a semigroup with a right identity and a right inverse for each element, where a semigroup is a nonempty set with a closed and associative binary operation. As a consequence it can be shown (proven) that a right inverse is also a left inverse, and it can be shown a right identity is also a left identity. Thus it is meaningful - after showing uniqueness of inverses and identities - to talk about _the_ two sided inverse and _the_ two sided identity. Typically the 'two sided' qualifier is suppressed and it is called _the_ inverse and _the_ identity.
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