Can Commutation between Subgroups be Achieved without Assuming Normality?

[mnb96]

Hello,
let's suppose I have two subgroups $R$ and $T$, and I know that in general they do not commute: that is, $rt\neq tr$ for some $r\in R$, $t\in T$.

Is it possible, perhaps after making specific assumptions on R and T, to find some $r'\in R$, and $t'\in T$ such that: $$rt=t'r'$$.

This is possible, for example, with some matrix manipulations if R and T are respectively the groups of rotations and translations in 2D. I was wondering if it is possible to find a more general algebraic approach without making explicit how R and T are defined.

 

[Bacle2]

Hi, mnb96. A few questions:

Do the subgroups have trivial intersection? Are they part of any specific supergroup?

How about looking at the group table (or, otherwise, how is the group given to you)?

 

[mnb96]

Hi Bacle2,
thanks for your help. I consider R and T as being subgroups of the supergroup G=RT, and I do not assume that R and T have trivial intersection.

However I noticed that if I assume that at least one of the two subgroups is normal, then I could solve the problem. Let's suppose for example that T is a normal subgroup of G=RT, then we have:
$$rt=[r,t]tr$$ where [r,t] is the commutator. Thus, $$rt=(rtr^{-1})t^{-1}tr$$ and since T is normal we have: $$rt=t'r$$ where $t'=rtr^{-1}\in T$.

If you define T as the group of 2D translations and R as the group of 2D rotations, and observe that T is a normal subgroup of G=RT, the above construction actually yields a well-known result...

I just wonder if it is possible to drop the assumption of normality and still come up with some more general result.