Construct a projection satisfying a certain property > Let $\cal G$ be a group of finite order $n$. For every prime divisor $p$ of $n$, construct a projection $P\in \cal N(G)$ such that $\operatorname{tr}_{\cal N(G)}(P)=1/p$. Here $\cal N(G)$ denotes the algebra of bounded, $G$-equivariant operators on $l^2(\cal G)$, and the trace of such an operator $T$ is defined by $$\operatorname{tr}_{\cal{N}(G)}(T)=\langle Te,e\rangle_{l^2(G)}$$ $e\in \cal G$ being the unit-element. Since $\cal G$ is finite, we can identify $\cal N(G)=\Bbb C G$ and under this identification, the trace of an element $\sum_g c_gg$ is just the coefficient of $e\in \cal G$. I'm struggling to write down a suitable element $P\in \Bbb C\cal G=\cal N(G)$ that has as its coefficient of the unit element $1/p$, and which satisfies $P=P^*=P^2$. I'm very thankful for any help.

If $g$ is an element of order $p$ (this always exists by Cauchy's Theorem), let $$ Q=\frac1p\,\sum_{j=0}^{p-1} g^j. $$ Then $\text{tr}(Q)=1/p$ and $$ Q^*=\frac1p\,\left(\sum_{j=0}^{p-1} g^j\right)^*=\frac1p\,\sum_{j=0}^{p-1} g^{-j} =\frac1p\,\sum_{j=0}^{p-1} g^{p-j}=\frac1p\,\sum_{k=1}^{p} g^k=\frac1p\,\sum_{k=0}^{p-1} g^k=Q. $$ Note that $$g(1+g+g^2+\cdots+g^{p-1})=g+g^2+\cdots+g^p=1+g+g^2+\cdots+g^{p-1}.$$ It follows that $$g^k(1+g+g^2+\cdots+g^{p-1})=1+g+g^2+\cdots+g^{p-1}$$ for all $k$. Then $$(1+g+g^2+\cdots+g^{p-1})^2=p\,(1+g+g^2+\cdots+g^{p-1}).$$ Thus, $Q^2=Q$.