Homomorphisms & Iso Theorems

G/ im .

The idea

Definition.

A homomorphism from a group $G$ to a group $H$ is a map $\varphi \colon G \to H$ with $\varphi(ab) = \varphi(a)\varphi(b)$ for all $a, b \in G$. Its kernel is $\ker\varphi = \{g \in G : \varphi(g) = e\}$ and its image is $\operatorname{im}\varphi = \{\varphi(g) : g \in G\}$.

Theorem (First Isomorphism Theorem).

Let $\varphi \colon G \to H$ be a homomorphism. Then $\ker\varphi$ is a normal subgroup of $G$, and $G/\ker\varphi \cong \operatorname{im}\varphi.$

A homomorphism need not be one-to-one: several elements of $G$ may map to the same element of $H$. The theorem describes the collapse exactly. Two elements $a$ and $b$ have the same image if and only if $a b^{-1}$ lies in the kernel, so $\varphi$ identifies precisely the elements within each coset of $\ker\varphi$ — the figure shows each coset $g\ker\varphi$ collapsing to the single point $\varphi(g)$ — and the quotient $G/\ker\varphi$ is an exact copy of the image.

Two consequences follow. For finite $G$, counting cosets gives $|G| = |\ker\varphi| \cdot |\operatorname{im}\varphi|$, so we compute the size of an image by dividing: $|\operatorname{im}\varphi| = |G| / |\ker\varphi|$. And $\varphi$ is injective if and only if $\ker\varphi = \{e\}$: a trivial kernel makes each coset a single element, so nothing is identified.

Ways to work on it

• Walkthrough. Kernels, the first isomorphism theorem, injectivity.
• Practice. Image size from the kernel.
• Hardest. Pin down every possible image size of a homomorphism between two cyclic groups.

Not sure where to start? Take the ten-question placement test.