Let $G=\operatorname{Sp}_n \mathbb{C}$ or $\operatorname{O}_n \mathbb{C}$, acting naturally on $V=\mathbb{C}^n$. Let $A$ be the subalgebra of $\operatorname{End}(V^{\otimes d}$) generated by $g \otimes … \otimes g$ for $g \in G$. In P.513 of Fulton and Harris' "Representation Theory A First Course", it is stated that by the simplicity of $G$, $A$ is a semisimple algebra. I am having trouble understanding why this is true.
Semisimple algebras are in fact not defined anywhere in the book, but to my understanding a (associative, not necessarily unital or commutative) algebra $A$ over a field is semisimple if it has no nontrivial nilpotent ideals. It can also be shown that this is equivalent to $A$ being a direct sum of simple algebras, a simple algebra being one without nontrivial proper ideals.
My first idea is to check that there are no nilpotent elements in $A$. Let $\lambda \in A$, $\lambda= \Sigma a_i (g_i \otimes … \otimes g_i)$ for some $a_i \in \mathbb{C}, g_i \in G$, but when one tries to compute $\lambda^k$ the cross terms in the product quickly become unmanageable.
Motivated by the author's remark about the simplicity of $G$, I also tried: suppose $I$ is an ideal in $A$, let $H=\{g \in G: g \otimes … \otimes g \in I\}$, I could check that $H$ is a normal subgroup of $G$. Hence $H=0$ if $I$ is proper, but then I'm stuck at showing $I$ cannot be nilpotent.
Any help would be appreciated!