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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?

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    $\begingroup$ If $\varphi(x) \in U$, then $\varphi(x^{-1}) = \varphi(x)^{-1} \in U$, since $U$ is a subgroup. $\endgroup$ Aug 26, 2013 at 12:42
  • $\begingroup$ @DanielFischer: I know. But this doesn't show that $\varphi^{-1}(U)$ has inverse. You only say that inverse elements in $G$ get mapped to inverse elements $H$. $\endgroup$ Aug 26, 2013 at 12:45
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    $\begingroup$ @moose If $a\in \varphi^{-1}(U)$, then $\varphi (a)\in U$, thus $\varphi (a)^{-1}=\varphi (a^{-1})\in U$ as $U$ is a group. But this means $a^{-1}\in \varphi^{-1}(U)$ (because its image is in $U$) and $\varphi^{-1}(U)$ has inverses. $\endgroup$
    – walcher
    Aug 26, 2013 at 12:50
  • $\begingroup$ And that means $x \in \varphi^{-1}(U) \Rightarrow x^{-1}\in \varphi^{-1}(U)$, so $K = \varphi^{-1}(U)$ is closed under inversion (and multiplication). $\endgroup$ Aug 26, 2013 at 12:50
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    $\begingroup$ @GerryMyerson: Ok, so I've posted the answer. But I made it community wiki, because the important ideas came from Daniel Fischer and walcher. $\endgroup$ Aug 26, 2013 at 13:09

5 Answers 5

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

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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

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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 $a$.

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Let $\pi: G \rightarrow H$ 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.

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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.

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