Kernels and Images

Every homomorphism comes with two subgroups: what it kills and what it reaches.

Definitions. For $\varphi: G \to H$, the kernel is $\ker\varphi = \{g \in G : \varphi(g) = e_{H}\}$ and the image is $\operatorname{im}\varphi = \{\varphi(g) : g \in G\}$.

Why it works

Theorem. (1) $\operatorname{im}\varphi \le H$. (2) $\ker\varphi$ is a normal subgroup of $G$. (3) $\varphi(a) = \varphi(b)$ exactly when $a^{-1}b \in \ker\varphi$; so $\varphi$ is one-to-one exactly when $\ker\varphi = \{e\}$. (4) If $G$ is finite, $|G| = |\ker\varphi| \cdot |\operatorname{im}\varphi|$.
Proofs. (1) $\varphi(a)\varphi(b)^{-1} = \varphi(ab^{-1})$ lies in the image. (2) If $\varphi(a) = \varphi(b) = e$, then $\varphi(ab^{-1}) = e$, so the kernel is a subgroup; and for any $g$, $\varphi(gag^{-1}) = \varphi(g)\,e\,\varphi(g)^{-1} = e$, so it is normal. (3) $\varphi(a) = \varphi(b) \iff \varphi(a)^{-1}\varphi(b) = e \iff \varphi(a^{-1}b) = e$. (4) By (3), the elements sent to $\varphi(a)$ form the coset $a\ker\varphi$; the cosets of the kernel match the points of the image one for one, so $[G : \ker\varphi] = |\operatorname{im}\varphi|$. $\square$

Worked example: Kernel and image in $\mathbb{Z}_{n}$

Find the kernel and image of $\varphi: \mathbb{Z}_{12} \to \mathbb{Z}_{18}$, $\varphi(x) = 3x \bmod 18$.

Well defined: $12 \cdot 3 = 36 \equiv 0 \pmod{18}$.
Kernel: $3x \equiv 0 \pmod{18}$ means $6 \mid x$, so $\ker\varphi = \{0, 6\}$.
Image: generated by $\varphi(1) = 3$: $\{0, 3, 6, 9, 12, 15\}$.
Check: $2 \cdot 6 = 12 = |\mathbb{Z}_{12}|$.

Answer: $\ker\varphi = \{0, 6\}$, $\operatorname{im}\varphi = \langle 3 \rangle$.

Worked example: Named kernels

Find the kernels of $\operatorname{sgn}: S_{4} \to \{\pm 1\}$ and $\det: GL_{2}(\mathbb{R}) \to \mathbb{R}^{*}$.

$\ker\operatorname{sgn}$ is the even permutations, $A_{4}$, with $12$ elements; the image $\{\pm 1\}$ has $2$, and $12 \cdot 2 = 24$.
$\ker\det$ is $SL_{2}(\mathbb{R})$, the matrices of determinant $1$; so $SL_{2}(\mathbb{R})$ is normal in $GL_{2}(\mathbb{R})$.

Answer: $A_{4}$ and $SL_{2}(\mathbb{R})$.

Worked example: Sizes

A homomorphism $\varphi$ maps a group of order $60$ with $|\ker\varphi| = 12$. How big is the image?

$|\operatorname{im}\varphi| = 60/12 = 5$.

Answer: $5$ elements.

Part (2) is also the fastest way to prove that a subgroup is normal: show that it is the kernel of some homomorphism. That is how we know at once that $A_{n}$ is normal in $S_{n}$, that $SL_{n}(\mathbb{R})$ is normal in $GL_{n}(\mathbb{R})$, and that $n\mathbb{Z}$ is normal in $\mathbb{Z}$, without conjugating a single element. Part (3) is the standard way to prove a homomorphism is one-to-one: compute the kernel instead of comparing pairs of elements.

Common mistake

Collecting the elements sent to the wrong identity. The kernel is everything sent to the identity of the target: that is $0$ for $\mathbb{Z}_{n}$ but $1$ for $\mathbb{R}^{*}$ or $\{\pm 1\}$. And the kernel always contains $e$ (by $\varphi(e) = e$), so it is never empty; "one-to-one" means it contains nothing else.

A shortcut for normality. To prove a subgroup is normal, it is often quickest to show it is a kernel. $A_{n}$ is normal in $S_{n}$ because it is the kernel of the sign map, and $SL_{n}(\mathbb{R})$ is normal in $GL_{n}(\mathbb{R})$ because it is the kernel of $\det$.

Key idea

$\ker\varphi$ (everything sent to $e$) is a normal subgroup; $\operatorname{im}\varphi$ is a subgroup; $\varphi$ is one-to-one iff the kernel is $\{e\}$; and $|G| = |\ker\varphi|\,|\operatorname{im}\varphi|$.

Sources and further reading

Thomas W. Judson, Abstract Algebra: Theory and Applications (open textbook), chapter 11, Homomorphisms (section 11.1)

John A. Beachy and William D. Blair, Abstract Algebra, online study guide (Northern Illinois University), section 3.7, Homomorphisms

MIT OpenCourseWare 18.703 Modern Algebra (Spring 2013), lecture on kernels and images

Kernels and Images

Practice

Builds on

Leads to

Sources and further reading