Prove these groups are not isomorphic

  • #1

Homework Statement



prove that R under addition is not isomorphic to[tex] R^*[/tex], the group of non zero real numbers under multiplication.

Homework Equations





The Attempt at a Solution



[tex]\varphi:(R,+) \rightarrow (R^* , .)[/tex]
let [tex]\varphi[/tex](x) = -x
then [tex]\varphi[/tex](x+y) = -(x+y) = -x-y
[tex]\neq \varphi(x)\varphi(y) = (-x)(-y) = xy \neq -x-y[/tex]
since these are not equal it proves they are not isomorphic?
im a little confused about this, am i doing this right?
Thanks all
 

Answers and Replies

  • #2
lanedance
Homework Helper
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2


how about considering multiplication by zero...

and for clarity, with
[tex]
\varphi:(R,+) \rightarrow (R^* , .)
[/tex]

your notation is not very clear, to clean it up a bit how about writing
[tex]
X = \varphi(x)
[/tex]

try re-writing the argument you made and see if it makes sense
 
  • #3
22,129
3,298


[tex]\varphi:(R,+) \rightarrow (R^* , .)[/tex]
let [tex]\varphi[/tex](x) = -x
then [tex]\varphi[/tex](x+y) = -(x+y) = -x-y
[tex]\neq \varphi(x)\varphi(y) = (-x)(-y) = xy \neq -x-y[/tex]
since these are not equal it proves they are not isomorphic?
im a little confused about this, am i doing this right?
Thanks all

This does not suffice to prove that they are not isomorphic. What you have done now, is prove that [tex]\varphi[/tex] is not an isomorphism. However, there could be another function, that does yield an isomorphism. It's not because one function is not an isomorphism, that there doesn't exist a function that is!!

You'll have to proceed by contradiction. Suppose that there does exist an isomorphism [tex]\varphi:(R,+)\rightarrow (R^*,.)[/tex] (and you know nothing else of this function, only that it's an isomorphism!), then try to derive a contradiction.
 
  • #4


thanks for the replys people,
so i know nothing else about this function only that it is isomophic, So the 2 groups are really the same only "labelled" differently. im not sure how to use this...
 
  • #5
22,129
3,298


So, since you know it is an isomorphism, then there must be an element x which gets sent to -1. But where does x+x gets sent to?
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,847
966


Homework Statement



prove that R under addition is not isomorphic to[tex] R^*[/tex], the group of non zero real numbers under multiplication.
Is this really what the problem says? I can't help but wonder about [itex]\phi(x)= e^x[/itex] where [itex]\phi[/itex] is from R to [itex]R^*[/itex]. Why is that not an isomorphism?
 
  • #7
Dick
Science Advisor
Homework Helper
26,263
619


Is this really what the problem says? I can't help but wonder about [itex]\phi(x)= e^x[/itex] where [itex]\phi[/itex] is from R to [itex]R^*[/itex]. Why is that not an isomorphism?

R* contains negative numbers. It's not onto.
 
  • #8


So, since you know it is an isomorphism, then there must be an element x which gets sent to -1. But where does x+x gets sent to?

im not sure if im thinking along the right lines but x+x get sent to x^2 where as -1-1 = -2 gets sent to +1???
 
  • #9
Dick
Science Advisor
Homework Helper
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im not sure if im thinking along the right lines but x+x get sent to x^2 where as -1-1 = -2 gets sent to +1???

Partially right. But confused. If phi(x)=(-1) then phi(x+x)=phi(x)*phi(x)=1. Now what does phi(0) have to be? 0 is the identity in R+, yes?
 
  • #10


Partially right. But confused. If phi(x)=(-1) then phi(x+x)=phi(x)*phi(x)=1. Now what does phi(0) have to be? 0 is the identity in R+, yes?

phi(0) = (0)
phi(0+0) = 0 in (R^*, .) but its supposed to be the group of non zero real numbers so we have a contradiction?
 
  • #11
22,129
3,298


Why would phi(0)=0??
 
  • #12


If phi(x)=(-1) then phi(x+x)=phi(x)*phi(x)=1

so phi(0)=(-1) then phi(0+0)=phi(0)*phi(0)=1

so phi(0) = (1)???
 
  • #13
Dick
Science Advisor
Homework Helper
26,263
619


If phi(x)=(-1) then phi(x+x)=phi(x)*phi(x)=1

so phi(0)=(-1) then phi(0+0)=phi(0)*phi(0)=1

so phi(0) = (1)???

phi(0)=1 all right. But your argument is hardly convincing since it relies on the statement phi(0)=(-1)??!! Where did that come from? Can't you really think of something a little more persuasive??
 

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