The First Isomorphism Theorem

A homomorphism collapses its kernel to the identity and keeps everything else apart. The first isomorphism theorem says that this collapse is exactly a quotient group.

Why it works

First isomorphism theorem. If $\varphi: G \to H$ is a homomorphism with kernel $K$, then $$G/K \cong \operatorname{im}\varphi, \quad \text{by } gK \mapsto \varphi(g).$$ Proof. Call the map $\psi$. Well defined: if $gK = g'K$ then $g^{-1}g' \in K$, so $\varphi(g)^{-1}\varphi(g') = \varphi(g^{-1}g') = e$ and $\varphi(g) = \varphi(g')$. Homomorphism: $\psi(gK\,g'K) = \psi(gg'K) = \varphi(gg') = \varphi(g)\varphi(g')$. One-to-one: $\psi(gK) = e$ means $g \in K$, so $gK = K$, the identity of $G/K$. Onto $\operatorname{im}\varphi$: every $\varphi(g)$ is $\psi(gK)$. $\square$

The theorem is a machine for naming quotients: to identify $G/N$, find a homomorphism out of $G$ whose kernel is $N$, and look at its image. It also explains the counting rule $|G| = |\ker\varphi|\,|\operatorname{im}\varphi|$.

Quotients named by the first isomorphism theorem (each map is onto its target).
Homomorphism	Kernel	Conclusion
$\mathbb{Z} \to \mathbb{Z}_{n}$, reduce mod $n$	$n\mathbb{Z}$	$\mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}_{n}$
$\operatorname{sgn}: S_{n} \to \{\pm 1\}$	$A_{n}$	$S_{n}/A_{n} \cong \mathbb{Z}_{2}$
$\det: GL_{n}(\mathbb{R}) \to \mathbb{R}^{*}$	$SL_{n}(\mathbb{R})$	$GL_{n}(\mathbb{R})/SL_{n}(\mathbb{R}) \cong \mathbb{R}^{*}$
$\mathbb{R}^{*} \to \mathbb{R}^{+}$, $x \mapsto \|x\|$	$\{\pm 1\}$	$\mathbb{R}^{*}/\{\pm 1\} \cong \mathbb{R}^{+}$

$\mathbb{Z}_{12}$ colored by the cosets of $\langle 4 \rangle$, with the table of the quotient. Reduction mod 4 sends each color to one element of $\mathbb{Z}_{4}$, so $\mathbb{Z}_{12}/\langle 4 \rangle \cong \mathbb{Z}_{4}$.

Worked example: A quotient of $\mathbb{Z}_{12}$

Identify $\mathbb{Z}_{12}/\langle 4 \rangle$.

$\varphi: \mathbb{Z}_{12} \to \mathbb{Z}_{4}$, $x \mapsto x \bmod 4$, is well defined since $4 \mid 12$, and it is a homomorphism onto $\mathbb{Z}_{4}$.
Its kernel is $\{0, 4, 8\} = \langle 4 \rangle$.
So $\mathbb{Z}_{12}/\langle 4 \rangle \cong \mathbb{Z}_{4}$.

Answer: $\mathbb{Z}_{4}$.

Worked example: Powers of $i$

Use $\varphi: \mathbb{Z} \to \mathbb{C}^{*}$, $\varphi(n) = i^{n}$, to identify a quotient of $\mathbb{Z}$.

$i^{m + n} = i^{m}i^{n}$, so $\varphi$ is a homomorphism.
Image $\{1, i, -1, -i\}$; kernel $\{n : i^{n} = 1\} = 4\mathbb{Z}$.
So $\mathbb{Z}/4\mathbb{Z} \cong \{1, i, -1, -i\}$.

Answer: $\mathbb{Z}/4\mathbb{Z} \cong \langle i \rangle$.

Worked example: Sizes from the theorem

$\varphi: \mathbb{Z}_{18} \to \mathbb{Z}_{12}$ has $\varphi(1) = 2$. Find $|\ker\varphi|$.

Well defined: $18 \cdot 2 = 36 \equiv 0 \pmod{12}$.
The image is $\langle 2 \rangle$ in $\mathbb{Z}_{12}$, of order $6$.
$\mathbb{Z}_{18}/\ker\varphi \cong \operatorname{im}\varphi$, so $|\ker\varphi| = 18/6 = 3$: the kernel is $\{0, 6, 12\}$.

Answer: $3$.

The recipe, used in each example above: (1) write down a map out of $G$; (2) check that it is a homomorphism; (3) find its image (often by showing it is onto); (4) compute its kernel and check that it is the subgroup you are dividing by. Then the theorem gives the isomorphism with no further work: no need to check well-definedness on cosets by hand.

Common mistake

Matching $G/\ker\varphi$ with the whole target $H$ when $\varphi$ is not onto. The theorem gives the image: for $\varphi: \mathbb{Z}_{18} \to \mathbb{Z}_{12}$ above, the quotient has $6$ elements, not $12$. And do not guess a quotient from its order alone: two groups of order 4 need not be isomorphic, so name the quotient through a map.

Key idea

$G/\ker\varphi \cong \operatorname{im}\varphi$ via $gK \mapsto \varphi(g)$. To name a quotient $G/N$, find a homomorphism with kernel $N$ and read off its image.

The First Isomorphism Theorem

Practice

Builds on

Leads to

Sources and further reading