Help with Direct Sums of Groups

  • #1

Homework Statement


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.


Homework 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


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:

Answers and Replies

  • #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]?
 
  • #3
jgens
Gold Member
1,583
50
The question is correct. Consider the exponential map.
 
  • #4
micromass
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
22,129
3,298
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]?

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.
 
  • #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:

Related Threads on Help with Direct Sums of Groups

Replies
2
Views
3K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
2
Views
253
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
2
Views
3K
Replies
3
Views
2K
Replies
9
Views
978
Replies
3
Views
3K
Replies
9
Views
902
  • Last Post
Replies
1
Views
953
Top