I am trying to understand unitary representations of compact groups $G$. Equip the class of all unitary irreducible representations of $G$ with the usual equivalence relation, unitary equivalence. We define the dual space as the set of equivalence classes with respect to this relation. The dual space cannot ever form a group. I would like to check that my reasoning as to why this is true is correct. This is my reasoning:
We can try to equip this set with binary operation $[\pi][\pi']:=[\pi\cdot\pi']$ where $(\pi\cdot\pi')(x)=\pi(x)\cdot\pi'(x)$. Then this operation is pointwise multiplication of irreducible representations (which are just group homomorphisms). $\textbf{But}$, irreducible representations of compact groups are finite dimensional therefore can be expressed as square matrices, e.g.
$\pi(x)=n\times n$ matrix, $\pi'(x)=m\times m$ matrix. Then pointwise multiplication (and indeed regular matrix multiplication) cannot be defined unless $n=m$. However, in general the dual space will have representations of different dimensions - thus we cannot define a sensible binary operation on them.
This is in contrast with the $\textit{dual group}$ for Abelian locally compact groups, because irreducibles are just continuous homomorphisms into $\mathbb{C}$ and can therefore be equipped with pointwise multiplication to give group structure.
$\textbf{One question}$ comes to mind for me: how do we know there are no other binary operations we might come up with, that would give the dual space a group structure?
