Let $\lambda$ be a partition. Then for sufficiently large $n$ there is a corresponding conjugacy class $K_\lambda(n)$ of $S_n$ (got by ignoring $1$'s in $\lambda$). Use $+$ to denote the sum of the elements in the group algebra ${\mathbb Q}S_n$, we may write $$K_\lambda(n)^+K_\mu(n)^+=\sum_\nu c_{\lambda,\mu}^\nu(n)K_\nu(n)^+,$$ where the $c_{\lambda,\mu}^\nu(n)$ are non-negative integers that depend on all $3$ partitions and on $n$. Say that each element of $K(\lambda)$ can be written as a product of $\ell(\lambda)$ transpositions, but no fewer. Throw away the $\nu$ for which $\ell(\nu)<\ell(\lambda)+\ell(\mu)$. Then the resulting $c_{\lambda,\mu}^\nu(n)=c_{\lambda,\mu}^\nu$ are independent of $n$ (H. K. Farahat, G. Higman, The centres of symmetric group rings, Proc. Royal Soc. London A 250 (1959) 212-221.).
Now let $H(t)=\prod_{i=1}^\infty(1-tx_i)^{-1}$ be the generating function for the complete symmetric functions. Suppose that $H(t)$ has Lagrange inverse $H^*(t)=\sum_{n=0}^\infty h_n^*t^n$. Then the corresponding symmetric functions $h_n^*$ are algebraically independent. Set $h_\lambda^*=\prod h_{\lambda_i}^*$.
Denote the dual (w.r.t. the usual symmetric bilinear form on symmetric functions) of $h_\lambda$ by $K_\lambda$, for each partition $\lambda$. Then Macdonald shows that the $K_\lambda$ form a basis for symmetric functions whose multiplication constants are: $$K_\lambda^+K_\mu^+=\sum_\nu c_{\lambda,\mu}^\nu K_\nu.$$