If $g \in S$, does this necessitate that $g^{\dagger} \in S$?

Question

In Lecture 5 of Aram Harrow's lecture series "Quantum Information Science II", he sets about proving that $\Pi_{S} = \frac{1}{|S|} \sum_{g \in S} g$ can be written as $\Pi_{S}= \prod_{i=1}^{l}(\frac{I + s_{i}}{2})$.

He observes that $g\Pi_{S}=\Pi_{S}$ for $g \in S$. He then uses this fact to state that if $g \in S$, then $g^{\dagger} \in S$ and therefore $g^{\dagger}\Pi_{S}=\Pi_{S}$ and therefore $\Pi_{S}^{\dagger}\Pi_{S} = \Pi_{S}$.

However, I do not understand how he came to the conclusion that if $g \in S$, then $g^{\dagger} \in S$. I know that because $S$ is a group, for every $g \in S$, there must exist an inverse $g^{-1} \in S$. However, this does not necessarily imply that $g^{\dagger} \in S$ because as far as I am aware, the stabilizer group is not necessarily unitary? It consists of the tensor products of unitary elements, but I am not aware of a fact that the tensor product of unitary operators are themselves unitary?

Answer 1

It is indeed the case that a tensor product of unitaries is a unitary. Let $$ U=\bigotimes_iU_i $$ Then $$ UU^\dagger=\bigotimes_i(U_iU_i^\dagger)=\bigotimes_iI=I, $$ so $U$ is unitary given that all the $U_i$ are.