Commutation between subgroups

  • Thread starter mnb96
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  • #1
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Main Question or Discussion Point

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

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

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.
 

Answers and Replies

  • #2
Bacle2
Science Advisor
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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)?
 
  • #3
710
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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:
[tex]rt=[r,t]tr[/tex] where [r,t] is the commutator. Thus, [tex]rt=(rtr^{-1})t^{-1}tr[/tex] and since T is normal we have: [tex]rt=t'r[/tex] where [itex]t'=rtr^{-1}\in T[/itex].

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.
 

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