Exercise 3.1.1 (Preimage of a normal subgroup)

Proof. Here $\phi :G\to H$ is a homomorphism and $E$ is a subgroup of $H$.

(a)

First $\phi (1_G)=1_H\in E$, thus $1_G\in {\phi }^{-1}(E)$, so ${\phi }^{-1}(E)\ne \varnothing$, and by definition, ${\phi }^{-1}(E)=\{x\in G\mid \phi (x)\in E\}\subseteq G$.

If $x\in {\phi }^{-1}(E)$ and $y\in {\phi }^{-1}(E)$, then $\phi (x)\in E$ and $\phi (y)\in E$. Since $E$ is a subgroup of $H$,

$$\phi (x{y}^{-1})=\phi (x)\phi {(y)}^{-1}\in E.$$

Therefore $x{y}^{-1}\in {\phi }^{-1}(E)$.

This shows that ${\phi }^{-1}(E)$ is a subgroup of $G$.

(b)

We suppose now that $E⊴H$. Let $n\in {\phi }^{-1}(E)$ and $g\in G$. Put $h=\phi (g)$. Then $h\in H$ and $\phi (n)\in E$. Since $E⊴H$, $$\phi (\mathit{gn}{g}^{-1})=\phi (g)\phi (n)\phi {(g)}^{-1}=\mathit{h\phi }(n)h^{-1}\in E,$$

therefore $\mathit{gn}{g}^{-1}\in {\phi }^{-1}(E)$. Since this is true for all $n\in N$ and all $g\in G$, this shows that

$${\phi }^{-1}(E)⊴G.$$

(c)

Consider the particular case where $E=\{1_H\}$.

Then $E⊴H$ and ${\phi }^{-1}(E)={\phi }^{-1}(\{1\})=\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$. By part (b),

$$\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )⊴G.$$