# Help with Direct Sums of Groups

thoughtinknot

## Homework Statement

Let $\mathbb{R}$*=$\mathbb{R}$\{0} with multiplication operation. Show that $\mathbb{R}$*=$\mathbb{I}$2 ⊕ $\mathbb{R}$, where the group operation in $\mathbb{R}$ is addition.

## Homework Equations

Let {A1,...,An}$\subseteq$A such that for all a$\in$A there exists a unique sequence {ak} such that a=a1+...+an where ak$\in$Ak for all k, then A=A1⊕...⊕An

## The Attempt at a Solution

Since $\mathbb{I}$2={-1,1} I don't think I can show that every a*$\in$$\mathbb{R}$* 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

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thoughtinknot
Am I wrong in thinking this question is incorrect since $\mathbb{R}$ is not contained in $\mathbb{R}$*, thus $\mathbb{R}$* ≠ $\mathbb{I}$2 ⊕ $\mathbb{R}$?

Gold Member
The question is correct. Consider the exponential map.

Staff Emeritus
Am I wrong in thinking this question is incorrect since $\mathbb{R}$ is not contained in $\mathbb{R}$*, thus $\mathbb{R}$* ≠ $\mathbb{I}$2 ⊕ $\mathbb{R}$?
So consider ($\mathbb{R}$+, x) the group of positive real numbers, where x is normal multiplication. Then there exists a mapping, exp:$\mathbb{R}$$\rightarrow$$\mathbb{R}$+ such that exp(r)=er.
This can easily be shown to be an isomorphism, then I can use the cyclic group $\mathbb{I}$2 to extend this isomorphism to the negative reals aswell.