$f: (G,*) \to (H,\cdot)$ set bijection, where $H$ is a group. Moreover, $f$ is such that $x*y$ iff $f(x*y)=f(x)\cdot f(y)$. Then $(G,*)$ is a group Let $G= \lbrace a \in R \mid a>0,a \neq 1 \rbrace$ and a binary operation defined as $a∗b:=a^{\log(b)}$ for every $a,b \in G$. Prove $G$ is a group
I can set a bijection from $f: G  \to \mathbb{R}- \lbrace 0 \rbrace $ as $f(a)=\log(a)$, and the inverse functions is given as $f^{-1}(a)= \operatorname{exp}(a)$ right? Someone asked this before here:
Is $(G,*)$ defined as $G=\lbrace a\in \mathbb{R} \: | \: a>0, \: a\neq 1 \rbrace$ and $a*b:=a^{log (b)}$ a group?
But I don't understand why this bijection and the fact $z=x*y$ if and only if $f(z)=f(x)\cdot f(y)$ shows $G$ is a group. I understand in this case $\log(a*b)=\log(a)\log(b)$, this means our map is a group homomorphism. But I cannot conclude from here that $G$ is a group, how this shows $G$ has an identity element, inverse element, and associativity?
 A: If you translate “one plus three” to German, compute the result in German, then translate back to English, you still get “four.” You aren’t really doing a different operation in German, and any properties of the operations in English are also properties of the operations in German, and visa versa.
This is an abstract but very similar case, with $f$ the “translation,” and the condition that $f(x*y)=f(x)\cdot f(y)$ indicating that the operations are “the same” in the two languages.
Since $\mathbb R^\times=(\mathbb R\setminus\{0\},\cdot)$ is a group, then we have the identity element is $f^{-1}(1)=e$ in $G.$
To find an inverse of $a\in G,$ compute:
$$b=f^{-1}(f(a)^{-1})$$
So to find the inverse of $a,$ we find the inverse in the corresponding element to $a$ in the known group, and then map back. We can write $b$ explicitly as $e^{1/\log a}$ and show that $a*b=e,$ but that obscures what the bijection is doing.
Instead, we have $$\begin{align}f(a*b)&=f(a)\cdot f(b)\\&=f(a)\cdot f(f^{-1}(f(a)^{-1}))\\&=f(a)\cdot f(a)^{-1}\\&=1\end{align}$$ The third step because $f(f^{-1}(z))=z.$ So $a*b=f^{-1}(1)=e.$
Similarly for associative operations.
What this really shows is that what we call the elements of a group doesn’t affect whether it is a group. A bijection $f:(A,*)\to (B,\otimes)$ such that $f(x*y)=f(x)\otimes f(y)$ means that if one is a group, the other is.
Proving it can be laborious, but if you think of it as “renaming” or “translating” elements, it becomes more intuitive.
