3.2 Images, inverse images and kernels

In this section, $G$ and $H$ are groups, and $f:G\to H$ is a group homomorphism. We write $e$ for the identity element of both $G$ and $H$ (and tell the difference by context).

Theorem 3.11

Suppose $J$ is a subgroup $G$ and $K$ is a subroup of $H$. Then

The image $f(J)$ is a subgroup of $H$.
The preimage $f^{-1} K$ is a subgroup of $G$.

Proof ▼

1: Since $e\in J$, $f(e)\in f(J)$. So $f(J)\neq \emptyset $, and we can use the one-step subgroup test. Take $y_1, y_2\in f(J)$. Then we can find $x_1, x_2\in J$ such that, for $i=1, 2$, $y_i = f(x_i)$. Then $y_1y_2^{-1} = f(x_1)f(x_2)^{-1} = f(x_1x_2^{-1}) \in f(J)$ as $x_1x_2^{-1}\in J$. So $f(J)\leq H$.

2: $f(e) = e\in K$ as $K\leq H$. So, again, we can use the one-step test. Suppose $x_1, x_2\in f^{-1}(K)$. Then, setting $y_i = f(x_i)$ for $i=1,2$, we get that $y_i\in K$ for $i=1,2$. But then $f(x_1x_2^{-1}) = f(x_1)f(x_2)^{-1} = y_1y_2^{-1}\in K$. So $x_1x_2^{-1}\in f^{-1}(K)$. And, therefore, by the one-step test, $f^{-1}(K)\leq G$.

Definition 3.12

The kernel $\ker f$ of $f$ is the preimage of the identity subgroup. In other words, $\ker f := f^{-1}(\{ e\} )$.

Example 3.13

Define a map $f:\mathbb {R}\to \operatorname{\mathrm{GL}}_2(\mathbb {R})$ by setting

\[ f(\theta ) := \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}. \]

Then, giving $\mathbb {R}$ and $\operatorname{\mathrm{GL}}_2(\mathbb {R})$ the obvious group structures (addition and matrix multiplication), $f$ is a group homomorphism. This follows from the angle addition formulas for $\cos $ and $\sin $.

We have

\[ \ker f = \{ \theta \in \mathbb {R}: \text{ $\cos \theta = 1$ and $\sin \theta = 0\} $}. \]

So $\ker f = 2\pi \mathbb {Z} = \{ 2\pi n: n\in \mathbb {Z}\} $.