Talk:Induced Group Product is Homomorphism iff Commutative

I disagree with your name change suggestion because $G$ itself does not need to be abelian, nor indeed specifically do $H_1$ or $H_2$ - just that all elements of $H_1$ commute with all elements of $H_2$. Whether or not this is possible without $H_1$ and $H_2$ themselves being abelian may be something else which needs to be established. --prime mover 10:43, 9 August 2012 (UTC)


 * I think I orignally destined that suggestion for the corollary; I must have glossed over the fact that there was a heading 'Corollary' in the first place. As for the counterexample to the general statement, consider products of non-abelian groups like $S_3 \times S_3$. They work. --Lord_Farin 10:48, 9 August 2012 (UTC)

Recent name change
... we've had this discussion -- the group is *not* necessarily Abelian, just that each element of $H_1$ commutes with each element of $H_2$. --prime mover (talk) 17:07, 27 September 2014 (UTC)