I've read that the wedge of two cyclic maps, $f\vee g$, does not need to be cyclic. Well, I understood the counter-example (see below) except by the fact that $S^1\vee S^1$ is not an H-space.
Where can I find a proof that $S^1\vee S^1$ is not an H-space?
C.E. Since $S^1$ is an H-space the identity map $1_{S^1}$ is cyclic. But $1_{S^1}\vee 1_{S^1}=1_{S^1\vee S^1}$ is not cyclic, for $S^1\vee S^1$ is not an H-space.
edited: I've found on Hatcher's book an exercise: Show that if $(X,e)$ is an H-space then $\pi_1(X,e)$ is abelian. Since we know that $\pi_1(S^1\vee S^1)=\mathbb{Z}\ast\mathbb{Z}$ is not abelian, the aim now is to solve the exercise.
The solution of the exercise cited above can be found on Hu's book Homotopy theory, Chapter III, section 11, Prop. 11.4.
Proposition 11.4. If $X$ is an H-space with $x_0$ as a homotopy unit, then the fundamental group $\pi_1(X,x_0)$ is abelian.