A homomorphism is a map between groups that respects the operations. You know several already: $\det$ turns matrix products into number products, $e^{x}$ turns sums into products, and the sign of a permutation turns composition into multiplication of $\pm 1$. Like linear maps between vector spaces, homomorphisms let us compare groups.
Definition. A map $\varphi: G \to H$ between groups is a homomorphism if $$\varphi(ab) = \varphi(a)\varphi(b) \quad \text{for all } a, b \in G,$$ where the product on the left is taken in $G$ and on the right in $H$. For example $\varphi: \mathbb{Z} \to \mathbb{Z}_{n}$, $x \mapsto x \bmod n$, and $\exp: (\mathbb{R}, +) \to (\mathbb{R}^{+}, \cdot)$, where the rule reads $e^{a + b} = e^{a}e^{b}$.
Why it works
(1) $\varphi(e_{G}) = e_{H}$: $\varphi(e) = \varphi(ee) = \varphi(e)\varphi(e)$, and cancelling $\varphi(e)$ gives $e_{H} = \varphi(e)$.
(2) $\varphi(a^{-1}) = \varphi(a)^{-1}$: $\varphi(a)\varphi(a^{-1}) = \varphi(aa^{-1}) = \varphi(e) = e$, and inverses are unique.
(3) $\varphi(a^{n}) = \varphi(a)^{n}$ for every integer $n$ (induction, with (2) for negatives).
(4) If $|a| = n$, then $|\varphi(a)|$ divides $n$: $\varphi(a)^{n} = \varphi(a^{n}) = \varphi(e) = e$. $\square$
Homomorphisms out of a cyclic group
If $G = \langle a \rangle$, then $\varphi(a^{k}) = \varphi(a)^{k}$, so $\varphi$ is decided by the single element $\varphi(a)$. For $\varphi: \mathbb{Z}_{m} \to \mathbb{Z}_{n}$ with $\varphi(1) = k$, the formula $\varphi(x) = kx \bmod n$ is well defined exactly when $mk \equiv 0 \pmod{n}$ (the elements $0$ and $m$ of $\mathbb{Z}$ are the same in $\mathbb{Z}_{m}$, so they must have the same image). Counting the solutions $k$ shows there are exactly $\gcd(m, n)$ homomorphisms $\mathbb{Z}_{m} \to \mathbb{Z}_{n}$.
Worked example: All homomorphisms $\mathbb{Z}_{12} \to \mathbb{Z}_{8}$
- Need $12k \equiv 0 \pmod{8}$, that is, $8 \mid 12k$, that is, $2 \mid 3k$, so $k$ is even.
- $k = 0, 2, 4, 6$: four homomorphisms, and $\gcd(12, 8) = 4$.
- For $k = 6$: $\varphi(5) = 30 \bmod 8 = 6$.
Answer: $\varphi(1) \in \{0, 2, 4, 6\}$.
Worked example: Homomorphism or not?
- (a) $\det(AB) = \det A \det B$: yes.
- (b) $\varphi(0) = 1$ is not the identity $0$: no.
- (c) $(xy)^{2} = x^{2}y^{2}$ because real multiplication commutes: yes.
Answer: (a) and (c).
Worked example: Counting homomorphisms into $S_{3}$
- A homomorphism is decided by $\varphi(1)$, and $\varphi(1)$ must have order dividing $6$.
- Every element of $S_{3}$ has order $1$, $2$ or $3$, all dividing $6$, and any such choice gives a well-defined map $k \mapsto \varphi(1)^{k}$.
- So each of the $6$ elements works: $6$ homomorphisms. Only the two choices of a 3-cycle give an image of size $3$; none is onto, since $S_{3}$ is not cyclic.
Answer: $6$.
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