Question regarding definition of isomorphism

Question

Here's my question:

Suppose $G_1$ and $G_2$ be groups and $f:G_1\to G_2$ be an isomorphism. Then can I conclude $f$ is bijective and for any $a,b \in G_1$, $f(ab)=f(a)f(b)$?

However my textbook defines a function $f:G_1\to G_2$ is an isomorphism if $f$ is bijective with the property that for any two elements $a$ and $b$ in $G_1$, $f(ab)=f(a)f(b)$.

However, is the converse necessarily true?

Answer 1

if you consider a map $f$ between groups $G,H$ you restrict naturally to those with the property $$f(ab)=f(a)f(b) \space \forall a,b \in G $$ A function $f$ of this kind is called a group homomorphism. It is a natural map between two groups, preserving the group structure.
Furthermore, if $f$ is injective, it is called a monomorphism (for those maps, we have $\text{ker}f=\{a\in G: f(a)=0_H\}=\{0_G\}$, if $f$ is surjective, an epimorphism and if $f$ is both injective and surjective, its an isomorphism.

Answer 2

Yes it is enough (despite being a silly definition).

1. $f(1_G) = 1_H$ :

indeed, let $h\in H$, there's $g \in G$ st $h = f(g)$. Hence: $$ h f(1_G) = f(g) f(1_G) = f(g 1_G) = f(g) = h \ \ \mathrm{and} \\ f(1_G) h = f(1_G) f(g) = f(1_G g) = f(g) = h$$

2. $\forall g \in G, \, (f(g))^{-1} =f(g^{-1}) $ :

Let $g \in G$, write $h:=f(g)$. We have $$h f(g^{-1}) = f(g) f(g^{-1}) = f(1_G) = 1_H$$ Hence, $f(g^{-1})$ is the (unique) inverse of $h$ : $h^{-1} = f(g^{-1})$.