Almost Commutative Property in Groups

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Homework Statement


I'm trying to figure out if the following property has a name:

for g\in G, h\in H, \exists h'\in H s.t. gh=h'g.

obviously this is not quite commutativity, but it seems like it might be useful in a variety of situations.

Homework Equations



I've just finished a proof that if a group K has two normal subgroups G and H, whose intersection is just the identity, and whose join is K, then there exists an isomorphism θ(g,h)=gh for all g in G and all h in H. The key to proving surjectivity involved the fact that since H is normal, ghg^{-1} is also in H (call this h') so h=g^{-1}h'g and

gh=gg^{-1}h'g=h'g

The Attempt at a Solution


I think I've seen this discussed elsewhere, just can't remember the name. ----commutativity?
 
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wnorman27 said:

Homework Statement


I'm trying to figure out if the following property has a name:

for g\in G, h\in H, \exists h'\in H s.t. gh=h'g.

obviously this is not quite commutativity, but it seems like it might be useful in a variety of situations.
For H a group, the "property" that there exists such an h' doesn't have a name because it is always true! For any such g and h, h'= ghg^{-1} must exist. And that is saying that h and h' are conjugates. Perhaps that is what you are looking for.

Homework Equations



I've just finished a proof that if a group K has two normal subgroups G and H, whose intersection is just the identity, and whose join is K, then there exists an isomorphism θ(g,h)=gh for all g in G and all h in H. The key to proving surjectivity involved the fact that since H is normal, ghg^{-1} is also in H (call this h') so h=g^{-1}h'g and

gh=gg^{-1}h'g=h'g

The Attempt at a Solution


I think I've seen this discussed elsewhere, just can't remember the name. ----commutativity?
 
I understand that in the context of groups, this h' is just the result of conjugation of h by g, but my thought was that perhaps this might occur outside of the context groups (say in cases where inverses may not exist)? I can't think of any examples of this though... maybe this is just not useful.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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