# Showing that preimage of a subgroup is a subgroup

## Homework Statement

Prove that if $f:G\to H$ is a group homomorphism and $K\leq H$ then the preimage of $K$, defined as $f^{-1}(K)=\{g\in G | f(g)\in K\}$, is a subgroup of $G$.

## The Attempt at a Solution

1) Note that $f^{-1}(K)$ is nonempty, since $f(e_G) = e_H$ and $e_H \in K$. So $e_G \in f^{-1}(K)$.

2) Let $x,y \in f^{-1}(K)$. So $f(x),f(y) \in K$ by the definition of the preimage of K. Then $f(xy^{-1}) = f(x)f(y^{-1}) = f(x)f(y)^{-1} \in K$ because $f(x),f(y) \in K$ and $K$ is closed under multiplication and inverses. Hence $xy^{-1} \in f^{-1}(K)$.

We conclude that $f^{-1}(K) \le G$.

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Mentor

## Homework Statement

Prove that if $f:G\to H$ is a group homomorphism and $K\leq H$ then the preimage of $K$, defined as $f^{-1}(K)=\{g\in G | f(g)\in K\}$, is a subgroup of $G$.

## The Attempt at a Solution

1) Note that $f^{-1}(K)$ is nonempty, since $f(e_G) = e_H$ and $e_H \in K$. So $e_G \in f^{-1}(K)$.

2) Let $x,y \in f^{-1}(K)$. So $f(x),f(y) \in K$ by the definition of the preimage of K. Then $f(xy^{-1}) = f(x)f(y^{-1}) = f(x)f(y)^{-1} \in K$ because $f(x),f(y) \in K$ and $K$ is closed under multiplication and inverses. Hence $xy^{-1} \in f^{-1}(K)$.

We conclude that $f^{-1}(K) \le G$.
Yes. Nothing to add this time.