t****b
发帖数: 482 | 1 如果G是一个可解群(solvable group), a, b是其中的两个二阶元素(ie, a^2=b^2=e,
identity)，那么ab有可能是几阶?
If G is a solvable group, and the orders of a and b are both 2, that is, a^2
=e=b^2, then what are the possible orders of ab? | l*****e
发帖数: 65 | 2
^2
I will try to give an answer, but really not sure about its correctness. So
use your own judgement to find out the truth.
My answer is: any number n. (b=a^{-1} for n=1, and we discuss n>1.)
The reason: build up an example of a solvable group for the number.
Given any n>=2, let H= the order n cyclic group. Consider an
involution y acting on H by y^{-1} x y = x^{-1}.
Let G be the group generated by semi-product of H by y. Namely,
G=
Then 1. G is solvable, as H is normal subgroup of index 2.
2. y and xy are both of order 2.
3. if a=y and b=xy, then ab=yxy=x^{-1} has order n.
【在 t****b 的大作中提到】
: 如果G是一个可解群(solvable group), a, b是其中的两个二阶元素(ie, a^2=b^2=e,
: identity)，那么ab有可能是几阶?
: If G is a solvable group, and the orders of a and b are both 2, that is, a^2
: =e=b^2, then what are the possible orders of ab?
| t****b
发帖数: 482 | 3 Thank you so much for your help. |
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