Is the pre-image of a subgroup under a homomorphism a group?

Let $a \in \varphi^{-1}(U)$ be any element in the pre-image of $U$. This means that $\varphi(a) \in U$. As $U$ is a group, it has an inverse $(\varphi(a))^{-1} \in U$. But homomorphisms are closed under inversion, so $(\varphi(a))^{-1} = \varphi(a^{-1}) \in U$. This means $a^{-1} \in \varphi^{-1}(U)$.

set $H=\phi^{-1}(U)$, then $H$ is a group iff $\forall a,b \in H$ we also have $ab^{-1} \in H$ i.e. $\phi(ab^{-1}) \in U$

suppose, then, that $\phi(a)=u \in U$ and $\phi(b)=v \in U$

since $\phi$ is a group-morphism, $\phi(ab^{-1}) = \phi(a) \phi(b^{-1}) = u \phi(b)^{-1} = uv^{-1} \in U$ as $U$ is a group