Following Spanier's book on algebraic topology chapter $1$, section $6$ about suspensions, I'm wondering about the following questions:
1) We know that $S^n$ is an $H$-cogroup for all $n\geq1$ because $S^n = S(S^{n-1})$ where $S$ is the "reduced suspension" functor, which has been proven to take values in the category of $H$-cogroups. Is it also true that $T^n = S^1 \times \dots \times S^1$ is an $H$-cogroup? If so, how does one show this?
2) Suppose the answer to the above question is yes (which I intuitively think it is), then is there an $H$-cogroup isomorphism $S^n \to T^n$? Or at least an $H$-cogroup homomorphism? If so, then there would be a natural transformation between the functors $\pi_{S^n}$ and $\pi_{T^n}$, or even a natural equivalence if the map $S^n \to T^n$ is isomorphic (but it cannot be a homeomorphism. I'm a bit confused here, can it they be $H$-cogroup isomorphic but not homeomorphic?)
3) What I'm really after is a correspondence between $\pi_{S^n}(X)$ and $\pi_{T^n}(X)$ for given $X$ and $n\geq1$. The two are group-isomorphic, right? Is the above the right way to show this? If not, what is?
Note on notation: $\pi_Q$ is a covariant functor from $\mathsf{hTop}_{\bullet}$ to $\mathsf{Grp}$ which sends $X\in \operatorname{Obj}(\mathsf{hTop}_{\bullet})$ to the homotopy classes of basepoint-preserving maps $\operatorname{hom}_{\mathsf{hTop}_{\bullet}}(Q,X)$ and a morphism $[f]\in \operatorname{hom}_{\mathsf{hTop}_{\bullet}}(X,Y)$ to $$([g]\mapsto[f\circ g])\in \operatorname{hom}_{\mathsf{hTop}_{\bullet}}(\operatorname{hom}_{\mathsf{hTop}_{\bullet}}(Q,X), \operatorname{hom}_{\mathsf{hTop}_{\bullet}}(Q,Y)).$$