Is the pre-image of a subgroup under a homomorphism a group? Let $(G, +)$ and $(H, \circ)$ be groups, $U$ a subgroup of $H$ and $\varphi: G \rightarrow H$ be a group homomorphism, i.e. 
$$\forall a, b \in G: \varphi(a+b) = \varphi(a) \circ \varphi(b)$$
Is the pre-image $\varphi^{-1}(U) = \{g \in G: \varphi(g) \in U\}$ also a group?
My try
$e_G \in \varphi^{-1}(U)$, because $\varphi(e_G) = e_H$ and $e_H \in U$.
Let $a, b \in \varphi^{-1}(U)$. Then $\exists x, y \in U: \varphi(a) = x$ and $\varphi(b) = y$. As $U$ is a group, $x \circ y \in U$. As $\varphi$ is a group homomorphism, $\underbrace{\varphi(a)}_x \circ \underbrace{\varphi(b)}_y = \varphi(a + b) \in U \Leftrightarrow (a+b) \in \varphi^{-1}(U)$.
But I still need to show: $\forall x \in \varphi^{-1}(U) \exists x^{-1} \in U: x \cdot x^{-1} = x^{-1} \cdot x = e_G$
How can I show this?

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
