pr.probability - Conditional expectation of convolution product equals..

Assuming $Omega$ has the structure for defining convolutions I don't think it is ever an algebra homomorphism. Take $X$ to be supported on $mathcal{G}^c$, i.e. take some set of non-zero measure in $mathcal{G}^c$ and let $X$ be a function whose support lies in that set, then $E(Xast Y|mathcal{G})neq 0$ but $E(X|mathcal{G})=0$ so $E(X|mathcal{G})ast E(Y|mathcal{G})=0$.

Edit: Scratch what I said. I was confusing sub-$sigma$-algebra with sub-algebra of random variables and even in the finite case my statement is completely incorrect. In almost every instance $E(X|mathcal{G})$ will not be zero as Jonas points out in the comments.