LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-02 (claude-opus-5)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced — the exact statement appears in a cited source; only wording and notation differ.
AI-adapted — a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated — a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A group homomorphism automatically satisfies $f(e) = e'$ and $f(g^{-1}) = f(g)^{-1}$ , and $f(g^{n}) = f(g)^{n}$ for every $n \in \mathbb{Z}$ ; for monoid homomorphisms preservation of the identity must be assumed

Statement

Let $G$ and $G'$ be groups with identities $e$ and $e'$ , and let $f : G \to G'$ be a group homomorphism (Monoid homomorphism and group homomorphism), so $f(xy) = f(x)f(y)$ for all $x, y \in G$ . Then:

$f(e) = e'$ ;
$f(g^{-1}) = f(g)^{-1}$ for every $g \in G$ ;
$f(g^{n}) = f(g)^{n}$ for every $g \in G$ and every $n \in \mathbb{Z}$ , powers being those of Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ .

For monoid homomorphisms the analogue of claim 1 is false, so preservation of the identity has to be part of the definition: the map $u : \mathbb{Z} \to \mathbb{Z}$ with $u(x) = 0$ for every $x$ satisfies $u(xy) = u(x)u(y)$ for the multiplicative monoid $(\mathbb{Z},\cdot,1)$ , yet $u(1) = 0 \ne 1$ .

Facts & Assumptions

Given: Groups $G$ , $G'$ with identities $e$ , $e'$ , and a function $f : G \to G'$ with $f(xy) = f(x)f(y)$ for all $x, y \in G$ (Monoid homomorphism and group homomorphism).

[A1]

$f(xy) = f(x)f(y)$ for all $x, y \in G$ .

[L1]

The group laws in $G$ and in $G'$ (Group and abelian group, Semigroup and monoid).

[L2]

Cancellation in $G'$ : $ab = ac$ implies $b = c$ , and $ba = ca$ implies $b = c$ (Cancellation in a group: $gx = gy$ or $xg = yg$ forces $x = y$ ; equivalently left and right translation by $g$ are bijections of $G$ , so $gx = h$ and $xg = h$ each have exactly one solution).

[L3]

Uniqueness of inverses in the sharp form: if $x$ is invertible and $yx = e'$ or $xy = e'$ , then $y = x^{-1}$ (In a monoid, a left inverse and a right inverse of the same element are equal; hence an invertible element has exactly one inverse, and it is two-sided); and $(u^{-1})^{-1} = u$ (In a group $e^{-1} = e$ , $(g^{-1})^{-1} = g$ and $(gh)^{-1} = h^{-1}g^{-1}$ , the order of the last product being essential).

[L4]

Powers: $g^{0} = e$ , $g^{\sigma(k)} = g^{k}g$ for $k \in \mathbb{N}$ , and $g^{x} = (g^{k})^{-1}$ when $x < 0$ and $-x = \iota(k)$ , where $\iota : \mathbb{N} \to \mathbb{Z}$ is the embedding of The naturals embed in the integers with image the nonnegative integers (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[L5]

$g^{-x} = (g^{x})^{-1}$ for every $x \in \mathbb{Z}$ (Exponent laws in a group: $g^{m+n} = g^{m}g^{n}$ and $(g^{m})^{n} = g^{mn}$ for all $m, n \in \mathbb{Z}$ , and $(gh)^{n} = g^{n}h^{n}$ when $g$ and $h$ commute).

[L6]

Induction on $\mathbb{N}$ (The principle of mathematical induction).

[L7]

$\mathbb{Z}$ is a totally ordered commutative ring, so exactly one of $0 \le x$ and $x < 0$ holds, and $x < 0$ gives $0 < -x$ (The integers form a commutative ring, The integers form a totally ordered ring, Arithmetic on the integers, Order on the integers, The integers as equivalence classes of pairs of naturals).

[L8]

$(\mathbb{Z},\cdot,1)$ is a commutative monoid: multiplication on $\mathbb{Z}$ is associative and commutative with $x \cdot 1 = x$ , and $x \cdot 0 = 0$ for every $x$ (The integers form a commutative ring, Arithmetic on the integers).

Proof

technique · direct

1.1

Applying [A1] with $x = y = e$ gives $f(e) = f(e\,e) = f(e)f(e)$ ; and $f(e) = e' f(e)$ because $e'$ is the identity of $G'$ .

A1L1

1.2

Applying [A1] with $y = g^{-1}$ gives $f(g)f(g^{-1}) = f(g g^{-1}) = f(e)$ .

A1L1

1.3

Base of claim 3 for natural exponents: $f(g^{0}) = f(e)$ and $f(g)^{0} = e'$ .

1.4

The monoid statement. In the commutative monoid $(\mathbb{Z},\cdot,1)$ the constant map $u(x) = 0$ satisfies $u(xy) = 0 = 0 \cdot 0 = u(x)u(y)$ for all $x, y$ , so it obeys the product law; but $u(1) = 0$ and the identity of the monoid is $1$ , and $0 \ne 1$ in $\mathbb{Z}$ . So a map obeying the product law between monoids need not send the identity to the identity, and (H2) is not redundant there.

L8given

2.1

From $e' f(e) = f(e)f(e)$ and cancellation in $G'$ we get $f(e) = e'$ , which is claim 1.

step 1.1L2

3.1

By steps 1.2 and 2.1, $f(g)f(g^{-1}) = e'$ , so $f(g^{-1})$ is a right inverse of the invertible element $f(g)$ , and uniqueness of inverses gives $f(g^{-1}) = f(g)^{-1}$ : claim 2.

step 1.2step 2.1L3

3.2

Claim 3 for natural exponents. The set of $k \in \mathbb{N}$ with $f(g^{k}) = f(g)^{k}$ contains $0$ , by step 1.3 and step 2.1, which give $f(g^{0}) = f(e) = e' = f(g)^{0}$ ; and it is closed under $\sigma$ , since $f(g^{\sigma(k)}) = f(g^{k} g) = f(g^{k}) f(g) = f(g)^{k} f(g) = f(g)^{\sigma(k)}$ . By induction it is all of $\mathbb{N}$ .

step 1.3step 2.1A1L4L6

4.1

Claim 3 for negative exponents. Let $x < 0$ and write $-x = \iota(k)$ with $k \in \mathbb{N}$ , possible since $0 < -x$ . Then $g^{x} = (g^{k})^{-1}$ , so $f(g^{x}) = f\bigl((g^{k})^{-1}\bigr) = f(g^{k})^{-1} = \bigl(f(g)^{k}\bigr)^{-1} = f(g)^{x}$ , the last equality being the definition of the negative power of $f(g)$ .

step 3.1step 3.2L4L5L7

5.1

Every integer is either nonnegative, hence of the form $\iota(k)$ and covered by step 3.2, or negative and covered by step 4.1; so claim 3 holds for every $n \in \mathbb{Z}$ .

step 3.2step 4.1L4L7

6.1

Claims 1, 2 and 3 are steps 2.1, 3.1 and 5.1, and step 1.4 shows the corresponding automatic identity preservation fails for monoids.

step 2.1step 3.1step 5.1step 1.4∎

Remarks

Cancellation is the whole difference. Step 2.1 turns $f(e)f(e) = e'f(e)$ into $f(e) = e'$ , and it can do so only because $G'$ is a group. In a monoid the element $f(e)$ is merely idempotent, and idempotents other than the identity exist, as $0$ in $(\mathbb{Z},\cdot)$ shows.
Claim 3 is what makes homomorphisms interact with orders. It gives $f(g)^{n} = e'$ whenever $g^{n} = e$ , so the order of $f(g)$ divides the order of $g$ when the latter is finite (If $\operatorname{ord}(g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n-1}$ are distinct, and $\langle g \rangle$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ). That consequence is used from the next page onwards.
Only one of the two one-sided equations is checked in step 3.1. That is enough: $f(g)$ is invertible in the group $G'$ , and for an invertible element a single one-sided equation identifies the inverse (In a monoid, a left inverse and a right inverse of the same element are equal; hence an invertible element has exactly one inverse, and it is two-sided).

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 57 results over 18 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Group homomorphism (Wikipedia) (standard reference, not scraped)
Monoid (Wikipedia) (standard reference, not scraped)

A group homomorphism automatically satisfies f(e)=e′f(e) = e'f(e)=e′ and f(g−1)=f(g)−1f(g^{-1}) = f(g)^{-1}f(g−1)=f(g)−1, and f(gn)=f(g)nf(g^{n}) = f(g)^{n}f(gn)=f(g)n for every n∈Zn \in \mathbb{Z}n∈Z; for monoid homomorphisms preservation of the identity must be assumed