Let G be a group,
a)Show that if a,b ∈ G, the the order of a in G is equal to the order of bab^-1 in G.
b)Suppose that a∈G is only element of order two in G, show that for all b ∈ G, ab=ba
How to find the order of a?
Like, a^16=e, what is the possible orders of element a in G?
Any suggestion help would really be appreciated!!
Let G be a group?
2015-10-06 2:37 am
回答 (2)
2015-10-06 2:42 am
✔ 最佳答案
SMASH IT WITH A HAMMER!!!
2015-10-06 4:58 am
For a) observe that bxb^-1byb^-1 = bxyb^-1 for any x, y and b in G. From this it follows by induction that
(bab^-1)^n = ba^nb^-1 for any n, and thus if N is the order of a, that (bab^-1)^N = ba^Nb^-1 = bb^-1 = e.
(I like using e for the identity element in a group, some people use 1.)
Now suppose (bab^-1)^k = e for some k < N, then e = bb^-1 = beb^-1 = b^-1(bab^-1)^k b = (b^-1 bab^-1 b)^k
by the same lemma with b and b^-1 reversing roles. But this last simplifies to a^k, which would contradict
the order of a being N, thus N is the smallest positive integer for which (bab^-1)^N = e, that is the order of bab^-1.
Part b follows from part a, but multiplying both sides on the right by b^-1. I leave you to fill in the details.
(bab^-1)^n = ba^nb^-1 for any n, and thus if N is the order of a, that (bab^-1)^N = ba^Nb^-1 = bb^-1 = e.
(I like using e for the identity element in a group, some people use 1.)
Now suppose (bab^-1)^k = e for some k < N, then e = bb^-1 = beb^-1 = b^-1(bab^-1)^k b = (b^-1 bab^-1 b)^k
by the same lemma with b and b^-1 reversing roles. But this last simplifies to a^k, which would contradict
the order of a being N, thus N is the smallest positive integer for which (bab^-1)^N = e, that is the order of bab^-1.
Part b follows from part a, but multiplying both sides on the right by b^-1. I leave you to fill in the details.
參考: I'm a mathematician. We try not to do peoples' math homework for them, which is why I left you the proof by induction and the details of part b to do yourself.
收錄日期: 2021-04-18 00:23:49
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