# 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. • Mr Davis 97