1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Help with Direct Sums of Groups

  1. Sep 27, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [itex]\mathbb{R}[/itex]*=[itex]\mathbb{R}[/itex]\{0} with multiplication operation. Show that [itex]\mathbb{R}[/itex]*=[itex]\mathbb{I}[/itex]2 ⊕ [itex]\mathbb{R}[/itex], where the group operation in [itex]\mathbb{R}[/itex] is addition.

    2. Relevant equations
    Let {A1,...,An}[itex]\subseteq[/itex]A such that for all a[itex]\in[/itex]A there exists a unique sequence {ak} such that a=a1+...+an where ak[itex]\in[/itex]Ak for all k, then A=A1⊕...⊕An

    3. The attempt at a solution
    Since [itex]\mathbb{I}[/itex]2={-1,1} I don't think I can show that every a*[itex]\in[/itex][itex]\mathbb{R}[/itex]* can be expressed in a unique way. For example let a+=a*+1 and a-=a*-1, then a*=a+-1=a-+1. Am I defining the cyclic group of order 2 wrong? I'm not that sure about direct sums, our prof spent 5 minutes on them and 40% of our assignment involves them :S
    Last edited: Sep 27, 2012
  2. jcsd
  3. Sep 28, 2012 #2
    Am I wrong in thinking this question is incorrect since [itex]\mathbb{R}[/itex] is not contained in [itex]\mathbb{R}[/itex]*, thus [itex]\mathbb{R}[/itex]* ≠ [itex]\mathbb{I}[/itex]2 ⊕ [itex]\mathbb{R}[/itex]?
  4. Sep 28, 2012 #3


    User Avatar
    Gold Member

    The question is correct. Consider the exponential map.
  5. Sep 28, 2012 #4
    Well, of course the question is incorrect. The sets can not be equal. However, what the question asks is not whether the sets are equal, but whether they are isomorphic. You need to find an isomorphism between the sets.
  6. Sep 28, 2012 #5
    Okay the exponential map...

    So consider ([itex]\mathbb{R}[/itex]+, x) the group of positive real numbers, where x is normal multiplication. Then there exists a mapping, exp:[itex]\mathbb{R}[/itex][itex]\rightarrow[/itex][itex]\mathbb{R}[/itex]+ such that exp(r)=er.

    This can easily be shown to be an isomorphism, then I can use the cyclic group [itex]\mathbb{I}[/itex]2 to extend this isomorphism to the negative reals aswell.
    Last edited: Sep 28, 2012
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook