Image

For a Homomorphism $\theta :G\to H$, the image is

$$\boxed{\mathrm{Im}(\theta )=\{\theta (g):g\in G\}\subseteq H.}$$

The set of all output values. Always a subgroup of $H$.

Why a Subgroup

By the Subgroup, take $h_1,h_2\in \mathrm{Im}(\theta )$, so $h_i=\theta (g_i)$ for some $g_i\in G$:

• $\theta (e_G)=e_H$, so $e_H\in \mathrm{Im}(\theta )$.
• $h_1{h}_2=\theta (g_1)\theta (g_2)=\theta (g_1{g}_2)\in \mathrm{Im}(\theta )$.
• $h_1^{-1}=\theta (g_1^{-1})\in \mathrm{Im}(\theta )$.

So $\mathrm{Im}(\theta )\le H$.

$\theta$ is Surjective iff $\mathrm{Im}(\theta )=H$

By definition.

In the First Isomorphism Theorem

$$G/\ker (\theta )\cong \mathrm{Im}(\theta ).$$

The quotient by the kernel matches the image. So image and ”$G$ modulo the kernel” carry the same information.

Examples

• $\det :{\mathrm{GL}}_n(\mathbb{R})\to {\mathbb{R}}^{\ast }$: image = ${\mathbb{R}}^{\ast }$ (every nonzero real is a determinant).
• Signature $\sigma :S_n\to \{\pm 1\}$ for $n\ge 2$: image = $\{\pm 1\}$.
• Inclusion $H\hookrightarrow G$: image = $H$.
• Trivial homomorphism $\theta (g)=e_H$: image = $\{e_H\}$, kernel = $G$.