Can the pre-image (under homomorphism) of a subgroup be empty?

Question

I'm asked to prove that if $E \leq H, \varphi^{-1}(E) \leq G$ where $\varphi: G \rightarrow H$ is an homomorphism.

I can show that $\varphi^{-1}(E)$ satisfies the group condition of $\forall x, y \in \varphi^{-1}(E), xy^{-1} \in \varphi^{-1}(E)$.

But how do I know that $\varphi^{-1}(E) \neq \emptyset$? I don't see why this needs to be true if $\varphi$ is not an isomorphism.

Answer 1

Think at the neutral element. You have:

• every subgroup contains the neutral element

• a homomorphism sends the neutral element of the domain-group $G$ to the neutral element of the range-group $H$.

Answer 2

Grouphomomorphism $\phi$ sends $\mathsf{id}_G$ to $\mathsf{id}_H\in E$ so $\mathsf{id}_G\in\phi^{-1}(E)$.

Answer 3

The group axioms imply that $E$ is non-empty, because it has a neutral element, which coincides with $\operatorname{id}_H\in E$. Because $\phi(\operatorname{id}_G)=\operatorname{id}_H$, we are done.