Conjugacy Classes
Not what you're looking for?
Let K={k1,....km} be a conjugacy class in the finite group G.
a) Prove that the element K=k1+k2+....km is the center of the group ring R[G]
(check that g^-1Kg=K for all gin G)
b) Let K1,....Kr be the conjugacy classes of G and for each Ki let Ki be the element of R[G] that is the sum of the members of Ki. Prove that an element alpha of R[G] is in the center of R[G] iff alpha=a1K1 +.....+arKr for some a1,...ar in R
Purchase this Solution
Solution Summary
Conjugacy classes are investigated.
Solution Preview
Proof:
a) Since K={k1,...,km} is a conjugacy class of a finite group G, then
for any g in G, gkig^(-1)=ki' is still in K, for any ki in K. And
K'={k1',...,km'} is a permutation of K={k1,...,km}.
So k1+...+km = k1'+...+km'. Let K=k1+...+km, K'=k1'+...+km',
then gKg^(-1)=K'=K. So gK=Kg.
Since G is finite, then G={g1,...,gn} and G is a basis of the group
ring R[G]. For any r and gi, we have rgiK=rKgi=Krgi.
Thus for any r=r1g1+...+rngn in R[G], we have
rK=(r1g1+...+rngn)K=r1g1K+...+rngnK=Kr1g1+...+Krngn
=K(r1g1+..+rngn)=Kr.
So K is in the center of R[G].
b) Since K1,...Kr are conjugacy classes in G, ...
Purchase this Solution
Free BrainMass Quizzes
Geometry - Real Life Application Problems
Understanding of how geometry applies to in real-world contexts
Graphs and Functions
This quiz helps you easily identify a function and test your understanding of ranges, domains , function inverses and transformations.
Solving quadratic inequalities
This quiz test you on how well you are familiar with solving quadratic inequalities.
Exponential Expressions
In this quiz, you will have a chance to practice basic terminology of exponential expressions and how to evaluate them.
Probability Quiz
Some questions on probability