# 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?

• If $\varphi(x) \in U$, then $\varphi(x^{-1}) = \varphi(x)^{-1} \in U$, since $U$ is a subgroup. Aug 26, 2013 at 12:42
• @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 $H$. Aug 26, 2013 at 12:45
• @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. Aug 26, 2013 at 12:50
• 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). Aug 26, 2013 at 12:50
• @GerryMyerson: Ok, so I've posted the answer. But I made it community wiki, because the important ideas came from Daniel Fischer and walcher. Aug 26, 2013 at 13:09

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

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 $a$.

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.

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.