Let K={k1,

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, from the above proof, we
know that K1,...,Kr are in the center of G, where Ki is also denoted
as the sum of all elements in Ki.
"<=": If alpha=a1K1+...+arKr, then for any x in R[G], we have
x*alpha=x(a1K1+...+arKr)=a1xK1+...+arxKr
=a1K1x+..+arKrx=(a1K1+...+arKr)x=alpha*x
Thus alpha is in the center of R[G].
"=>": If alpha is in the center of R[G], then alpha is a linear combination
of elements in G. But elements in G is divided into conjuacy classes.
So alpha=K1'+K2'+...+Kr', where Ki' is a linear combination of elements
in Ki. For each g in G, g can be also considered an elements in R[G],
gKi'g^(-1) is also a linear combination of elements in Ki'. But alpha
is in the center of G, then g*alpha*g^(-1)=alpha. Thus we must have
gKi'g^(-1)=Ki.
Without the loss the generality, we only consider K1'(assume K1' is not 0)
Suppose h1,...,hm are all elements in K1 and K1'=b1h1+...+bmhm.
The trivial case is that K1 contains only one element, then K1'=b1h1=b1K1
and we are done. For nontrivial case that K1 contains at least two elements.
First I claim the bi is not 0 for any 1<=i<=m. If bi=0, then we can find
some g in G, such that ghjg^(-1)=hi, and bj is not 0, for some j not equal to i.
So gK1'g^(-1) contains item gbjhjg^(-1)=bjhi but K1' does not contain the term
hi because bi=0. So gK1'g^(-1) is not equal to K1'. This is a contradiction.
Second, I claim the b1=b2=...=br. Because if any two are not equal, say
bi is not equal to bj, then ghjg^(-1)=hi and gK1'g^(-1) has term bjhi, but
K1' only has term bihi and bi not equal to bj. So gK1'g^(-1) is not equal to K1'.
Again we get a contradiction.
Therefore, K1'=b1(h1+...+hm)=b1K1. Other Ki's are the same.
Thus alpha=a1b1K1+...+arbrKr and we are done.