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The kernel of a group homomorphism is a normal subgroup


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.1 Exercise 3.1.1

Let $\varphi : G \rightarrow H$ be a group homomorphism and let $E \leq H$ be a subgroup. Prove that $\varphi^*[E]$ (the $\varphi$-preimage of $E$) is a subgroup of $G$. If $E \trianglelefteq H$, prove that $\varphi^*[E] \trianglelefteq G$. Deduce that $\mathsf{ker}\ \varphi \trianglelefteq G$.


Solution: Let $x,y \in \varphi^*[E]$. Then $\varphi(x)$, $\varphi(y) \in E$, and since $E$ is a subgroup, $$\varphi(x)\varphi(y)^{-1} = \varphi(xy^{-1}) \in E$$ by the subgroup criterion. So $xy^{-1} \in \varphi^*[E]$, and by the subgroup criterion we have $\varphi^*[E] \leq G$.

Suppose $E$ is normal in $H$. Now let $g \in G$ and $x \in \varphi^*[E]$; we have $$\varphi(gxg^{-1}) = \varphi(g)\varphi(x)\varphi(g)^{-1} \in \varphi(g) E \varphi(g)^{-1} = E,$$ since $E$ is normal. Thus $gxg^{-1} \in \varphi^*[E]$, hence $\varphi^*[E]$ is normal in $G$.

The trivial subgroup $1 \leq G$ is trivially normal for all $G$, and by definition $\mathsf{ker}\ \varphi = \varphi^*[1]$. So $\mathsf{ker}\ \varphi$ is normal in $G$.


Linearity

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