Help with Direct Sums of Groups

[thoughtinknot]

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,...,A_n}[itex]\subseteq[/itex]A such that for all a[itex]\in[/itex]A there exists a unique sequence {a_k} such that a=a₁+...+a_n where a_k[itex]\in[/itex]A_k for all k, then A=A₁⊕...⊕A_n

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

 

[thoughtinknot]

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]?

 

[jgens]

The question is correct. Consider the exponential map.

 

[micromass]

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.

 

[thoughtinknot]

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)=e^r.

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.