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Problem 77 (The Kernel Of A Homomorphism Is A Subgroup)

Suppose that $f:G\to H$ is a homomorphism and that $e_H$ is the identity in $H$. Prove that the kernel of $f$, namely $\ker f$ or $f^{-1}(e_H)$, is a subgroup of $G$.

Solution

In order to prove that $\ker f$ of a group $G$ is a subgroup of $G$, we will use the subgroup test. Recall that $$\ker(f)=f^{-1}(e_H) = \{g\in G\mid f(g)=e_H\}.$$ Also recall that the subgroup test says that if $G$ is a group, if $\ker f$ is a nonempty subset of $G$, if $a,b \in \ker f$ implies $ab \in \ker f$, and if $a \in \ker(f)$ implies $a^{-1} \in \ker f$, then $\ker f$ is a subgroup. By assumption, we know that $G$ is a group.

We will now show that $\ker{f}$ is a nonempty subgroup of $G$. Let $g \in \ker(f)$. By definition of $\ker(f)$, we know that $g \in G$. Thus, we know that $\ker(f) \subseteq G$. Additionally, consider the identity in $G$, call it $e_G$. From Problem 76, we know that $f(e_G) = e_H$. Thus, we know $e_g \in \ker(f)$. Because $e_G \in \ker{f}$ and $\ker{f} \subseteq G$, we know that $\ker{f}$ is a nonempty subset of $G$.

We will now show that $\ker{f}$ is closed under taking multiplications. Let $a,b \in \ker(f)$. We compute $$f(ab) = f(a)f(b) = e_H.$$ Thus, $ab \in \ker(f)$.

We will now show that $\ker{f}$ is closed under taking inverses. Let $a \in \ker(f)$. We know that $a^{-1}$ is in $G$. Recall that we know that $f(a^{-1}) = f(a)^{-1}$ from Problem 76. We compute $$f(a^{-1}) = f(a)^{-1} = e_H^{-1} = e_H.$$ Thus, we know that $a^{-1} \in \ker{f}$. It follows that $\ker{f} \leq G$.

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