A confusion on $\Omega$ and $\Sigma$ functors We know that for any pointed topological space there are two functors
$$
\Omega:\left\{\text{pointed topological spaces}\right\}\longrightarrow \left\{\text{H-groups}\right\}
$$
and
$$
\Sigma:\left\{\text{pointed topological spaces}\right\}\longrightarrow \left\{\text{H-cogroups}\right\}
$$
which are adjoint functors for each other.
Moreover, we have
$$
[X,\Omega^{n}Y]\simeq[\Sigma X,\Omega^{n-1}Y]=[\Sigma X,\Omega(\Omega^{n-2}Y)]
$$
where $Y$ is a pointed topological space with base point $y_0$. What confused me is, why is $[\Sigma X,\Omega(\Omega^{n-2}Y)]$ an abelian group? If it is, then we can conclude that all homotopy groups with $n>1$ are abelian.
I would appreciate it if you could help me!
 A: For each pointed space $Z$ we have the following five maps:

*

*Diagonal map $\Delta : Z \to Z \times Z$


*Folding map $\phi : Z \vee Z \to Z$


*Inclusion map $\iota : Z \vee Z \to Z \times Z$


*Coordinate flip map on product: $T : Z \times Z \to Z \times Z$


*Coordinate flip map on wedge : $T' : Z \vee Z \to Z \vee Z$
Note that $T \circ \Delta = \Delta$, $\phi \circ T' = \phi$ and $T \circ \iota
 = \iota \circ T'$.
An $H$-group $G$ has a multiplication map $\mu : G \times G \to G$ such that $[\mu] \circ [\iota] = [\phi]$ (here $[-]$ denotes pointed homotopy class). This reflects the fact that multiplication has a two-sided identity. Here we do not care about associativity and existence of inverses.
Dually, an $H$-cogroup $K$ has a comultiplication map $\kappa : K \to K \vee K$ such that $[\iota] \circ [\kappa] = [\Delta]$.
If $G$ is an  $H$-group and $K$ an $H$-cogroup, then the set $[K,G]$ of pointed homotopy classes gets two multiplications, one induced from $\mu$ (call it $\cdot$) and one induced from $\kappa$ (call it $*$). Both multiplications agree. To see this, look at the following diagram:
$\require{AMScd}$
\begin{CD}
K  @>{\kappa}>> K \vee K  @>{f \vee g}>> G \vee G @>{\phi}>> G \\
@V{id_K}VV @V{\iota}VV @V{\iota}VV @V{id_G}VV \\
K @>{\Delta}>> K \times K @>{f \times g}>> G \times G @>{\mu}>> G
\end{CD}
It commutes up to homotopy, the bottom row represents $[f] \cdot [g]$ and top row represents $[f] * [g]$.
Let us now check that multiplication in $[K,G]$ is commutative:
$$[g] \cdot [f] = [\mu \circ (g \times f)\circ \Delta] = [\mu \circ(g \times f)\circ T\circ \Delta] = [\mu \circ T \circ (f \times g)\circ \Delta] \\= [\mu \circ T \circ \iota \circ (f \vee g)\circ \kappa] = [\mu \circ \iota \circ T' \circ (f \vee g)\circ\kappa] = [\phi \circ T' \circ (f \vee g)\circ\kappa] \\= [\phi \circ (f \vee g)\circ\kappa] = [f] * [g] = [f] \cdot [g].$$
Intuitively, the reason for commutativity is that the coordinate flip is "absorbed" both by $\Delta$ and $\phi$.
This shows in particular that $[\Sigma X, \Omega Y]$ is always an abelian group.
