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?
 A: 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
A: Why not just use the 1 Step Subgroup Test? Please notify me if my attempt here is defectible?
$\phi^{-1}(U)$ is a subgroup iff $\forall a,b \in \phi^{-1}(U)$,  $ab^{-1} \in \phi^{-1}(U)$. 
By definition of preimage, this means $\phi(ab^{-1}) \in U.$
$\iff \phi(a)\phi(b^{-1}) \in U$.
This last line is true. Here's why. $b \in \phi^{-1}(U) \iff \phi(b) \in U \iff$ Its inverse is in U $ \iff (\phi(b))^{-1} = \phi(b^{-1}) \qquad \in U$. Same logic with $a$.
A: 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 $a^{-1} \in \varphi^{-1}(U)$.
A: Let $\phi: G \rightarrow  H$ be a group homomorphism. Let $E \leq H$ be a subgroup. We show the set
$$\pi^{-1}(E)=\{g \in G : \pi(g) \in E\}$$
is a subgroup of $G$. Its enough to show if $h_1,h_2 \in \pi^{-1}(E)$ that we have $h_1h_2^{-1} \in \pi^{-1}(E).$ We check:
\begin{align}
\pi(h_1h_2^{-1})&= \pi(h_1)\pi(h_2)^{-1} && \text{as $\phi$ is a group homomorphism}\\
& \in E && \text{as $E$ is closed under the operations and as $h_1,h_2 \in \pi^{-1}(E)$}
\end{align}
Thus $h_1h_2^{-1} \in \pi^{-1}(E)$ and we have $\pi^{-1}(E) \leq G$ as needed.
A: Just a comment, if $U$ is normal in $H$, then $\varphi^{-1}(U) = \ker(G \to H \to H/U)$ which is then a normal subgroup.
