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show/hide this revision's text 2 typos and deleted my doubts, as the comments overcame my doubts

I looked at my homotopy theory lecture notes and we had the following similar result: $X$ H-CoGroup, $Y$ $H$-Group, then both group structures defined of on [X,Y] agree. The proof goes roughly as follows: Call the upper products $\cdot$, resp. $*$. Inserting the definitions of thos those products, one can show , the following "distributivity"distributivity":

$(a\cdot b)*(c\cdot d)=(a * c)\cdot(b * d)$

Then one shows that both products have the same neutral element and finally

$f*g=(f\cdot 1) * (1\cdot g)=(f * 1)\cdot(1 * g)=f\cdot g$,

gives the result. That's the strategy of the proof in the case of $H$-(co-)groups.I don't know, whether the upper (weaker) assumptions also allow a proof like this.
show/hide this revision's text 1

I looked at my homotopy theory lecture notes and we had the following similar result: $X$ H-CoGroup, $Y$ $H$-Group, then both group structures defined of [X,Y] agree. The proof goes roughly as follows: Call the upper products $\cdot$, resp. $*$. Inserting the definitions of thos products, one can show, the following "distributivity"

$(a\cdot b)*(c\cdot d)=(a * c)\cdot(b * d)$

Then one shows that both products have the same neutral element and finally

$f*g=(f\cdot 1) * (1\cdot g)=(f * 1)\cdot(1 * g)=f\cdot g$,

gives the result. That's the strategy of the proof in the case of $H$-(co-)groups. I don't know, whether the upper (weaker) assumptions also allow a proof like this.