Prove: $\Phi^{-1}(e_2)$ is a subgroup of $G_1$, where $e_2$ is the unit element of $G_2$. [closed]

Suppose $(G_1, \circ)$ and $(G_2, *)$ are both groups:

$\Phi: G1\to G2$, $\Phi$ is surjective and $\forall a,b \in G_1$ has:

$$\Phi(a\circ b)= \Phi(a)* \Phi(b)$$

Prove: $\Phi^{-1}(e_2)$ is a subgroup of $G_1$, where $e_2$ is the unit element of $G_2$.

Solution 1:

$\phi : G_1 \to G_2 $ be an homomorphism.

Then, \begin{align} \phi^{-1}(e_2)&=\{g\in G : \phi(g)\in \{e_2\} \} \\ &=Ker(\phi)\end{align}

To show, $Ker(\phi) \le G_1$

Choose, $a, b\in Ker(\phi) $

Then, $\phi(ab^{-1} ) = \phi(a)\phi(b)^{-1}=e_2$

Hence, $ab^{-1} \in Ker(\phi) $

This implies $\phi^{-1}(e_2) \le G_1$

Not only subgroup , it is a normal subgroup of $G_1$ and the proof is not difficult.

Note: $\phi $ need not be onto homomorphism to the above proof. Since, any homomorphism from $G_1$ to $G_2$ map $e_{G_1}$ to $e_{G_2}$.