What is exactly meant by "preserves the group operation"? I'm new to the topic and I'm reading "Contemporary Abstract Algebra" by Gallian an this is how isomorphism is defined:
"An isomorphism $\phi$ from a group $G$ to a group $\overline{G}$ is a one-to-one mapping (or function) from $G$ onto $\overline{G}$ that preserves the group operation."
What is exactly meant by "preserves the group operation"? The one of $G$?  $\overline{G}$?
 A: Let $g\cdot h$ denote the multiplication in $G$ and $\overline g*\overline h$ be multiplication in $\overline G$.  Then a homomorphism $f:G\to\overline G$ is a map such that
$$f(g\cdot h)=f(g)*f(h).$$
A: It means that if $(G, \ast_{G},e_{G})$ and $(\overline{G}, \ast_{\overline{G}},e_{\overline{G}})$ are groups with their respective operation $\ast_{G}$ and $\ast_{\overline{G}}$. Then $\phi: G \to \overline{G}$ is an isomorphism if $\phi$ is bijective and for every $x,y \in G$:
$$\phi(x\ast_{G}y)=\phi(x)\ast_{\overline{G}}\phi(y).$$
The latter condition means that $\phi$ preserves the group operations.
A: If you have two elements $g_1, g_2 \in G$, there are two ways to produce an element in $\bar{G}$:


*

*use the binary operation $\ast$ of $G$ to produce the element $g_1\ast g_2$ of $G$, then apply the map $\phi$ to obtain $\phi(g_1\ast g_2) \in \bar{G}$.

*apply the map $\phi$ to both $g_1$ and $g_2$ to obtain $\phi(g_1), \phi(g_2) \in \bar{G}$, then use the binary operation $\cdot$ of $\bar{G}$ to produce the element $\phi(g_1)\cdot\phi(g_2) \in \bar{G}$.


The map $\phi$ is said to preserve the binary operation if it doesn't matter which of the above two processes we use. That is, $\phi$ satisfies $\phi(g_1\ast g_2) = \phi(g_1)\cdot\phi(g_2)$ for every $g_1, g_2 \in G$. In some sense, $\phi$ maps 'products' in $G$ to 'products' in $\bar{G}$.

Here is a more abstract way of viewing the above condition using the notion of transport of structure.
For convenience, let $b : G\times G \to G$ and $\bar{b} : \bar{G}\times\bar{G} \to \bar{G}$ denote the binary operations on $G$ and $\bar{G}$ respectively. As we have a bijection $\phi : G \to \bar{G}$ we can 'pullback' the binary operation $\bar{b}$ to a binary operation $\phi^*\bar{b}$ on $G$ as follows: for $g_1, g_2 \in G$, we define 
$$\phi^*\bar{b}(g_1, g_2) := \phi^{-1}(\bar{b}(\phi(g_1), \phi(g_2))).$$ 
Now we have two binary operations on $G$, namely $b$ and $\phi^*\bar{b}$. The condition above is equivalent to the statement that $\phi^*\bar{b} = b$ (i.e. the two binary operations are the same). 
In the special case where $\bar{G} = G$, we have $\bar{b} = b$ so the condition reads $\phi^*b = b$. That is, the group operation of $G$ is preserved by (the pullback via) $\phi$.
A: It means that $f(e_G)= e_{\overline G}$, and for all $x,y\in G, f(xy) = f(x)f(y)$, which together imply that $f(x^{-1}) = f(x)^{-1}$.
