LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-02 (claude-opus-5)

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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.

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A group homomorphism automatically satisfies $f (e) = e^{'}$ and $f (g^{- 1}) = f (g)^{- 1}$ , and $f (g^{n}) = f (g)^{n}$ for every $n \in Z$ ; 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 $f (x y) = f (x) f (y)$ 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 $n \in Z$ , 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 $u : Z \to Z$ with $u (x) = 0$ for every $x$ satisfies $u (x y) = u (x) u (y)$ for the multiplicative monoid $(Z, \cdot, 1)$ , yet $u (1) = 0 \neq 1$ .

Facts & Assumptions

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

[A1]

$f (x y) = f (x) f (y)$ 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 $G^{'}$ : $a b = a c$ implies $b = c$ , and $b a = c a$ implies $b = c$ (Cancellation in a group: $g x = g y$ or $x g = y g$ forces $x = y$ ; equivalently left and right translation by $g$ are bijections of $G$ , so $g x = h$ and $x g = h$ each have exactly one solution).

[L3]

Uniqueness of inverses in the sharp form: if $x$ is invertible and $y x = e^{'}$ or $x y = e^{'}$ , 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 $(g h)^{- 1} = h^{- 1} g^{- 1}$ , the order of the last product being essential).

[L4]

Powers: $g^{0} = e$ , $g^{σ (k)} = g^{k} g$ for $k \in N$ , and $g^{x} = (g^{k})^{- 1}$ when $x < 0$ and $- x = ι (k)$ , where $ι : N \to Z$ 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 $x \in Z$ (Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute).

[L6]

Induction on $N$ (The principle of mathematical induction).

[L7]

$Z$ is a totally ordered commutative ring, so exactly one of $0 \leq x$ 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]

$(Z, \cdot, 1)$ is a commutative monoid: multiplication on $Z$ 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

1.1A1L1

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 $G^{'}$ .

1.2A1L1

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

1.3L4

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

1.4L8given

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

2.1step 1.1L2

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

3.1step 1.2step 2.1L3

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.

3.2step 1.3step 2.1A1L4L6

Claim 3 for natural exponents. The set of $k \in N$ 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 $σ$ , since $f (g^{σ (k)}) = f (g^{k} g) = f (g^{k}) f (g) = f (g)^{k} f (g) = f (g)^{σ (k)}$ . By induction it is all of $N$ .

4.1step 3.1step 3.2L4L5L7

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

5.1step 3.2step 4.1L4L7

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

6.1step 2.1step 3.1step 5.1step 1.4∎

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.

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 $(Z, \cdot)$ 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 $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ 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).

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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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