Groups: Order of an element divides the order of the group

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An application of Lagrange's theorem
  1. Adam Glesser
  2. Group-theory
  3. Lagrange
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Комментарии
Автор

Oh, it's rather quite simple but it has eluded me all day. Thanks a lot, that makes sense.

aiiishiba
Автор

So, typically, the element x will not generate the entire group G. However, x will always generate the subgroup (x) = {x^i | i an integer}, and the order of x equals the order of (x). By Lagrange's theorem the order of (x) divides the order of G, and so the order of x must divide the order of G.

AdamGlesser
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Dude you're a G, this helped me out so much fam

lilhonky
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Thanks a lot for all the lectures on Math especially on " COMPLEX ANALYSIS " . Can you please upload a lecture series on " GROUP THEORY " .
Again Lots and lots and lots of thanks .

tawabullas
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What if the element doesn't generate the group, how do you prove it divides the element of the group.

aiiishiba
Автор

I would love to, and in the future might. First, I need to get a grant!

AdamGlesser