Group Theory Proof: If g^n = e then the order of g divides n

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I've purchased two of your courses on Udemy as supplemental instruction for this semester (with me already retaking Real Analysis), and let me tell you how much your videos have significantly improved my understanding of these concepts. I'm a concrete math kind of person, so the hypotheticals and theoreticals can sometimes confuse me, but you break it down into an amazing digestible format. Thank you. Seriously.

missjdiva

I think making it explicit that 0 is neither positive nor negative would help clarify the end, I think a lot of people are going to think of 0 as positive for some reason (that's how my mind classifies it)

MrCoreyTexas

Nice way to prove it, but its already intuitive.
If o(g) = n
then g^n = e (definition)
thus for g^m = e, -> m must be a multiple of n (because e*e= e)

Ex:
(g^n) * (g^n) * (g^n) = e*e*e = e = g^(3n)
It *has* to be a multiple of n

ptyamin

I just discovered your channel. Thanks for all your videos :) I don't understand one thing. n is not equal to the order of g ?

awazin

I dont understand.isnt n alreday order of g since g to the n is the identity.what is this o(g) u referring to here?

MariaPaula-fcmh

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