# Endomorphisms in a symmetric monoidal category

Let $\mathcal{C}$ be a symmetric monoidal category generated by one element $X$ such that $End(X)=G$ where $G$ is a finite group. Is it true that, for any object $A \in \mathcal{C}$, $End(A)$ is isomorphic to a wreath product $G \wr S_n$, $n \in \mathbb{N}$ ?

-

No. Take the category whose objects are vector spaces of dimension $2^n$ over $\mathbb{F}_2$ for $n \in \mathbb{Z}_{\ge 0}$ and whose morphisms are all isomorphisms. The object $X = \mathbb{F}_2^2$ generates this category under tensor product (I assume this is what you meant) and $\text{End}(X) \cong \text{GL}_2(\mathbb{F}_2) \cong S_3$. It should be straightforward to verify that $\text{End}(X^{\otimes n}) \cong \text{GL}_{2^n}(\mathbb{F}_2)$ has order larger than $S_3 \wr S_n$ for sufficiently large $n$.
Thank you for your answer. Could you detail the last part ? I have trouble following "generated by $\mathbb{F}_2^2$"... –  AlexPof Jun 19 '12 at 6:07
this is a really nice answer ! Is there a general way to derive $End(X)$ in a symmetric monoidal category given $G$ ? –  AlexPof Jun 19 '12 at 16:14
@Alex: you mean $\text{End}(X^{\otimes n})$? All you can say in general is that it admits a morphism from $\text{End}(X) \wr S_n$ (in a precise sense; in the symmetric monoidal category freely generated by $X$ and only the morphisms $X \to X$ there should be no other endomorphisms of $X^{\otimes n}$). –  Qiaochu Yuan Jun 19 '12 at 22:12
Yes, sorry, I meant $\text{End}(X^{\otimes n})$. If I understand correctly, the fact that $\text{End}(X^{\otimes n})$ in the example you gave is bigger than $\text{End}(X) \wr S_n$ comes from the fact that $X$ has more morphisms in the most general case ? The statement in my question was indeed meant in the context of a symmetric monoidal category freely generated by X with no other morphisms $X \to X$ than those of $G$... –  AlexPof Jun 20 '12 at 5:50