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Commutation between subgroups

  1. Aug 27, 2012 #1
    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.
  2. jcsd
  3. Aug 27, 2012 #2


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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)?
  4. Aug 27, 2012 #3
    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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