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?
Best Answer
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 $a^{-1} \in \varphi^{-1}(U)$.