Question

Question:Prove that the additive group$\mathrm{ℝ}$of all real numbers is not isomorphic to the multiplicative group ${\mathrm{ℝ}}^{*}$or nonzero real numbers.

if there were an isomorphism $f:\mathrm{ℝ}\to {\mathrm{ℝ}}^{*}$,then$f\left(k\right)=-1$ for some k.

use this fact to arrive at acontradiction.

Short answer

Answer

Additive group $\mathrm{ℝ}$of all real number is not isomorphic to multiplicative group${\mathrm{ℝ}}^{*}$.

Step-by-step solution

Referring to the Hint.

As given in the question,

$\mathrm{ℝ}=$additive group of all real numbers.

${\mathrm{ℝ}}^{*}=$multiplicative group of all non-zero numbers

Assuming an isomorphism$f:\mathrm{ℝ}\to {\mathrm{ℝ}}^{*}$ ,

Given by, f (k)

Proving that the additive group of all real numbers is not isomorphic to the multiplicative group ℝ* 

As we know,$\mathrm{ℝ}$ is a group with respect to addition. So, if $K\in \mathrm{ℝ}$,and $K\ne 0$the nth additive power of kis $nk,(n\in N)$. Hence, it can be seen that all the elements in has$\mathrm{ℝ}$ order 1.

As${\mathrm{ℝ}}^{*}$is a multiplicative group,$K\in \mathrm{ℝ}$ so if , then the nth multiplicative power of kwill be. Now, if we take some ,f (k) = -1

Then,(-1)²=1

So, the order of elements${\mathrm{ℝ}}^{*}$of is 2.

Since order of element of$\mathrm{ℝ}$ is 1 and order of element ${\mathrm{ℝ}}^{*}$in is 2, no elements in can produce elements of${\mathrm{ℝ}}^{*}$ . Thus,$f:\mathrm{ℝ}\to {\mathrm{ℝ}}^{*}$ does not hold for both the groups and hence, they are not isomorphic.