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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

We must show that the kernel of $f$ is nonempty, that it is a subset of $H$, that it is closed under the group operation, and that it is closed under inverses.

Let $e_G$ be the identity of $G$. Then $f(e_G) = e_H$ where $e_H$ is the identity in $H$, so $\ker(f)$ is non-empty. We know that $\ker(f)$ is a subset of $H$ by definition of the kernel.

Now we will show that $\ker(f)$ is closed under the group operation. Let $a, b \in \ker(f)$. Then $f(a) = f(b) = e_H$. Since $f$ is a homomorphism, we know that $f(ab) = f(a)f(b)$. This means that $f(ab) = e_He_H = e_H$, so we know that $ab\in\ker(f)$.

Last, we show that $\ker(f)$ is closed under taking inverses. Let $a\in \ker(f)$. Let $b$ be the inverse of $a$ in $G$. Then we have that $ab=e_G$, so $f(ab) = e_H$. Since we know that $f(a)=e_H$ and $f$ is a homomorphism, we know $e_H = f(b)e_H$. Thus we have that $f(b) = e_H$, so $f(b)$ is in $\ker(f)$.

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