Show that if g^l = 1, then the order of the group divides l
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Show that if g^l = 1, then the order of the group divides l

[From: ] [author: ] [Date: 12-04-12] [Hit: ]
b are natural numbers, a -Suppose k does not divide l.Then by ordinary division l = m*k +rwhere m is an integer >=0 and r is an integer >0.So1 = g^l= g^(m*k+r) = g^(m*k)* g^r but g^k = 1 so g^(m*k) =1.I strongly suggest that you erase this proof and see if you can do it yourself now.Thats how you learn how to do proofs.......
This is for a first-year intro to proofs class at a Canadian Uni:
Let G be a group, where k is a natural number and g is an element of the group s.t k is the smallest natural number s.t. g^k = 1
Show that:
if g^l = 1, then k divides l
if a,b are natural numbers, a < b, s.t. g^a = g^b, then show that k divides (b-a)

-
Suppose k does not divide l.
Then by ordinary division l = m*k +r where m is an integer >=0 and r is an integer >0.
So 1 = g^l= g^(m*k+r) = g^(m*k)* g^r but g^k = 1 so g^(m*k) =1.
Now we have 1 =g^r where r
I strongly suggest that you erase this proof and see if you can do it yourself now.
That's how you learn how to do proofs.
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