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 $\operatorname{im}\varphi = \{\varphi(g) : g \in G\}$.

Por qué funciona

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 $[G : \ker\varphi] = |\operatorname{im}\varphi|$. $\square$

Ejemplo resuelto: 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: $2 \cdot 6 = 12 = |\mathbb{Z}_{12}|$.

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

Ejemplo resuelto: 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 $GL_{2}(\mathbb{R})$.

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

Ejemplo resuelto: Sizes

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

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

Respuesta: $5$ 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.

Error común

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 $\det$.

Idea clave

$\ker\varphi$ (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 $|G| = |\ker\varphi|\,|\operatorname{im}\varphi|$.

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