According to wiki, Isomorphism:
Given two groups $(G, ∗)$ and $(H, {\displaystyle \odot })$, a group isomorphism from $(G, ∗)$ to $(H, {\displaystyle \odot })$ is a bijective group homomorphism from $G$ to $H$. Spelled out, this means that a group isomorphism is a bijective function ${\displaystyle f:G\rightarrow H}$ such that for all $u$ and $v$ in $G$ it holds that
$${\displaystyle f(u*v)=f(u)\odot f(v)}$$
Somewhere, I found this example:
Consider the group $G$ of real numbers under multiplication.
Let, $$G'=G \space \space \text{and} \space \space f(x)=2x \space\space\space\space \forall x \in G$$ Now, $f: G→G'$ is an isomorphism
Since, $$f(x_1*x_2)=\color{red}{f(2(x_1*x_2))=f(2(x_1\times x_2))=2\times2(x_1\times x_2)}=4x_1x_2$$
Now, $$f(x_1)=2x_1 \space\space f(x_2)=2x_2 →f(x_1)*f(x_2)=4x_1*x_2$$
Therefore, $$f(x_1*x_2)=f(x_1)*f(x_2)$$ It can be easily verified that $f(x) = 2x$ is both one to one and onto function from $G$ to $G'$.
Hence, $f(x)=2x$ is an isomorphism from $G$ to $G'$.
I understood except red colored line:
How we are taking $2$ and replacing $*$ to $\times$ operator. Again, taking $2$, after removing $f$. Could you please explain it ?
Related question:
Also notice that $f(x) = x^3$ is only a homomorphisms, but not isomorphism from $G$ to $G'$, since it is neither one to one nor onto from $G$ to $G'$.
In this $f(x) = x^3$, if it is neither one to one (i.e., monomorphism) nor onto (i.e., epicmorphism) from $G$ to $G'$, so can not be bijection (i.e., isomorphims); then,
How is it even Homomorphims ? Could you please explain it? Could you please provide proof that $f(x) = x^3$ is neither one to one nor onto from $G$ to $G'$.
f is not an isomorphism of the multiplicative group of reals because
2×3×5 = f(3×5), while f(3)×(5) = 2×3×2×5 = 2×f(3×5).
Exercise. Show f is an isomorphism of the additive group of reals