# showing that groups are isomorphic

• August 30th 2009, 10:19 PM
alexandrabel90
showing that groups are isomorphic
I have to show that the function Ø gives the isomorphism of G with G', by showing that Ø is a one to one function and is onto G'.

From my notes,
(1) if xØ = yØ, then e^x= e^y, so x=y. thus Ø is one to one.
(2) if r in R+, then
(In r)Ø = e ^ (In r) = r
where ( In r ) in R. thus Ø is onto R+.

why does (1) and (2) prove that Ø is one to one and is onto?

thanks!
• August 30th 2009, 10:36 PM
Sampras
To prove that a function is one to one, you show that $f(x) = f(y) \Rightarrow x = y$ or $x \neq y \Rightarrow f(x) \neq f(y)$. Then to prove surjectivity, you want to show that every element in the codomain has a pre-image. For example, suppose $f(x) = x^2$ and we define it from $\{1,2,3 \}$ to $\{1,4,9 \}$. 1 has preimage 1, 4 has preimage 2, and 9 has preimage of 3. So it is surjective.