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The sum of two odd integers plus one is odd, so the operation is closed. Both associations of a triple give $a+b+c+2$, proving associativity. Its identity element is $-1$, and the inverse element of $a$ is $-a-2$, also odd, since $a+(-a-2)+1=-1$. Thus it is an abelian group.

More explicitly, shifted addition on odd integers is ordinary integer addition in different coordinates:This bijection is a group isomorphism to $(\mathbb{Z},+)$.

Lagrange's theorem states that for a subgroup $H$ of a finite group $G$,The left cosets of $H$ partition $G$, and multiplication by a coset representative is a bijection from $H$ onto that coset, explaining the formula. If $|G|=p$ is prime, then $|H|$ is either $1$ or $p$, so $H=\{e\}$ or $H=G$. Thus the correct conclusion is no proper nontrivial subgroup. The trivial subgroup $\{e\}$ is proper for every group of prime order, so the wording in the question needs this qualification.

A homeomorphism $f:M\to N$ is a bijection for which both $f$ and $f^{-1}$ are continuous maps, using the respective metric topologies.Continuity of a bijection alone is insufficient: continuity of the inverse is a separate requirement.

The map $h(z)=e^{\pi z/2}$ is a conformal bijection from the width-two strip to the upper half-plane, and $h(i)=i$. Its inverse there is $h^{-1}(w)=2L(w)/\pi$.

Write $e^{\pi \beta /2}=A+iB$, where $B>0$. Conjugate the preceding affine half-plane map by $h$:Every argument of $L$ lies in the upper half-plane. The formula is therefore a conformal bijection of the strip onto itself, andThe final equality uses $0<\mathrm{Im}(\pi \beta /2)<\pi$, so no multiple of $2\pi i$ is lost. This is a strip automorphism with a prescribed image of its midpoint.

Write $\alpha =a+ib$ with $b>0$. ThenIt preserves the upper half-plane, has inverse $(z-a)/b$, and sends $i$ to $\alpha$. Its nonzero derivative $b$ verifies that this bijection is conformal.

In a countable first-order model, the finitely many named constants split the order into the open intervalstogether with the $n$ named points. Each interval is a countable dense linear order without endpoints in its induced order. It is nonempty and infinite by density and the absence of endpoints.

Use Cantor's classification of countable dense linear orders without endpoints: each such interval is isomorphic to the rational order. It follows by a back-and-forth method extending finite increasing partial bijections alternately along enumerations of the two orders. Given two first-order models of ${\mathrm{DLO}}_n$, choose isomorphisms between their corresponding intervals and send each named point to its counterpart. Their union is an increasing bijection preserving all constants, hence an isomorphism in $L_n$.

There are countable first-order models, for example any finite increasing list in the rational order. ThereforeFor $n=0$, the same argument is just the classification theorem for the whole order.

Every automorphism preserves all logical formulas by induction on their construction, so an automorphism taking $\vec{a}$ to $\vec{b}$ necessarily gives equal partial types.

For the converse, first note that a finite first-order structure is atomic by (ii). Thus in both cases every realized finite-tuple partial type has an isolating logical formula. Suppose two finite tuples $\vec{u},\vec{v}$ have the same complete type, and choose a new element $x$ to extend the first tuple. Let $\theta (\vec{z},w)$ isolate the realized partial type of $(\vec{u},x)$. Sincelogical equality of partial types gives $\mathcal{A}\models \exists w\theta (\vec{v},w)$. Choose a witness $y$. Isolation ensures $(\vec{u},x)$ and $(\vec{v},y)$ have the same complete type. In particular, if $x$ is distinct from the entries already used, their inequality logical formulas ensure $y$ is distinct from its corresponding entries. The resulting finite map is injective and partial elementary.

Start from $\vec{a}\mapsto \vec{b}$, which is well-defined even for repeated coordinates because their logical equality patterns agree. Enumerate the carrier and extend alternately to include the next missing element in the domain and the next missing element in the range. For a range step apply the same extension argument to the inverse finite map. The union is an elementary bijection from $A$ onto itself: every finite tuple and every logical formula involving it is handled at some stage. It is therefore an automorphism taking $\vec{a}$ to $\vec{b}$. In the finite case the construction terminates once all elements have been included.

HenceCountability makes the back-and-forth numeralwise enumeration in arithmetic exhaust the entire first-order structure; it cannot simply be dropped.

Use the first-order language with one equivalence relation $E$. Let $\mathcal{A}$ have exactly two equivalence classes, one countably infinite and one uncountable, with no further first-order structure. Its complete theory says that there are exactly two classes and both are infinite.

This first-order theory has quantifier elimination. To eliminate one existential logical variable from a conjunction of atomic formulas and negated atomic formulas, a logical equality either fixes its value, or finitely many inequalities exclude finitely many points. Its required equivalence-class membership is determined by the finitely many existing classes represented among the parameters. A required unused class exists exactly when fewer than two classes have already been used. Once its class is possible, infinitude supplies a point avoiding the finite exclusions. These tests are quantifier-free logical equality and equivalence conditions on the parameters. Disjunctive normal form and induction eliminate all quantifiers and also show completeness.

Thus a finite tuple's complete type is isolated by its finite equality-and-equivalence pattern. Every realized finite-tuple partial type is principal, so $\mathcal{A}$ is atomic. In particular all individual elements have the same complete one-type, isolated by $x=x$.

Choose $a$ in the countable class and $b$ in the uncountable class. Any automorphism taking $a$ to $b$ would give a bijection from their whole equivalence classes, impossible because their cardinalities differ. ThereforeIts uncountability is precisely what avoids the conclusion of (iii).

Take the first-order theory of an infinite set in the first-order language of logical equality alone, with an axiom asserting at least $n$ distinct elements for each positive integer $n$. It is complete and eliminates quantifiers: a new logical variable either equals a specified old logical variable or is required to avoid only finitely many existing values, which an infinite domain permits.

For every infinite cardinal $\kappa$, any two first-order models of size $\kappa$ are isomorphic because every bijection between their underlying sets preserves logical equality, the whole first-order language. ThusIts complete $n$-types record only the partition of coordinate indices according to logical equality. There are finitely many such partitions, and each is isolated by the corresponding logical equalities and inequalities. Ryll-Nardzewski therefore recovers countable categoricity; there are no nonprincipal complete finite-arity partial types to omit in this example.

Use the first-order theory $T_{\mathbb{Z}}$ of a bijection $f$ with named inverse $g$ and no finite cycles. Its axioms areEvery first-order model is a disjoint union of free two-sided integer orbits. Write $f^k$ for positive powers of $f$ and for powers of $g$ when $k$ is negative.

This first-order theory is complete and eliminates quantifiers. Every first-order term reduces to $f^k(x_i)$, and atomic logical formulas reduce to statements $x_j=f^k(x_i)$. To eliminate an existential logical variable in a conjunction of literals, first remove self-equalities using the no-finite-cycle axioms. If an equation fixes that logical variable as an integer translate of another logical variable, substitute that value. If no such equation remains, only finitely many points are excluded by the remaining inequalities, so some value exists in the infinite domain. Apply this to disjunctive normal form and then induct on logical formulas. The resulting parameter-free first-order sentences are decided, proving completeness.

For each fixed finite $n$, a complete type records a partition of the logical variables into common orbits and finitely many integer offsets within each block. There are countably many possibilities. Countable first-order models with one orbit and with two orbits are not isomorphic, so the first-order theory is not ${\aleph }_0$-categorical.

Thus this example has the source's explicitly defined countable-type property. In fact it also satisfies ordinary omega-stability: over a countable parameter set, an element is either a specified integer translate of a parameter, giving countably many possibilities, or lies in an entirely new orbit, giving one further complete one-type. The same finite-orbit-pattern description handles finite tuples. ThereforeThe source prints the label “$0$-stable”; the construction meets its stated definition as well as the usual stronger omega-stability condition. The difference between countably many and finitely many partial types is exactly why Ryll-Nardzewski does not yield categoricity here.

Every vector space has a basis and is classified up to linear isomorphism by its dimension. Over a finite field, a vector is a finite linear combination of basis Elements. If the basis has infinite cardinality $\lambda$, the set of finite such combinations has cardinality $\lambda$: there are at least $\lambda$ basis vectors and at most a countable union of finite powers of $\lambda$, each of size $\lambda$. Thus the underlying cardinality of an infinite space equals its dimension.

Any two models of cardinality $\kappa \ge {\aleph }_0$ therefore have bases of cardinality $\kappa$. A bijection between their bases extends uniquely to an $L$-isomorphism. A space with that basis size exists, giving the required existence clause as well. Hence $T$ is categorical in every infinite cardinal.

The cardinality of a vector space is central: here “infinite” must continue to refer to the underlying set. Over a countably infinite field, every nonzero finite-dimensional space is already infinite. Thus the model class now consists of all nonzero $K$-vector spaces, not only infinite-dimensional ones.

Since independence of a pair is not definable over an infinite field, for (i), independence of a single vector is still $x\ne 0$, but independence of tuples of length at least two is not definable by one formula in this one-sorted language. Quantifier elimination follows from the linear-equation argument in the preceding part for all these infinite domains. Any candidate two-variable formula therefore involves only finitely many linear equations. Take an independent pair, and compare it with $(x,{\lambda }_{a}x)$, where $x\ne 0$ and $a\in K$ avoids the finitely many ratios making one of those nontrivial equations vanish. Both pairs satisfy the same listed atomic formulas, contradicting a proposed definition of independence. Longer tuple independence would define pair independence by existentially completing a pair inside an infinite-dimensional model, giving the same obstruction.

For (ii), the vector-space axiom schema together with $\exists x(x\ne 0)$ axiomatizes the infinite underlying spaces. For (iii), there is again no universal axiomatization, because the zero subspace is a substructure.

For finite axiomatizability with infinitely many scalar symbols, in (iv) there is no finite axiomatization in the full language: any finite axiom set mentions only finitely many scalar symbols. In a nonzero vector space alter an unused nonzero ${\lambda }_a$ to send zero to a nonzero vector, retaining all mentioned interpretations. The finite sentences keep their truth values, but a vector-space axiom is now false. Relative to the already imposed vector-space axioms, the extra condition is finitely axiomatized by the single nonzero-vector sentence. These absolute and relative answers must be distinguished.

For (v), the one-variable algebra still has two atoms and four Elements. In two variables the isolated atoms are $(0,0)$ and the countably many rank-one directions, represented by $x\ne 0,y={\lambda }_{a}x$ and by $x=0,y\ne 0$. There is also a nonisolated complete type of an independent pair. A formula using finitely many linear equations either picks finitely many isolated atoms, or contains that independent type and all but finitely many isolated atoms. Therefore $F_2/E_2(T)$ is the finite-cofinite Boolean algebra on a countably infinite set of atoms, with the independent type the extra nonisolated point in its Stone space. In particular this algebra is countably infinite, rather than the full power set of its atoms.

For (vi), dimension can be $1,2,3,\dots$ or ${\aleph }_0$ while the domain is countable, producing pairwise nonisomorphic countable models. Thus $T$ is not countably categorical. For an uncountable $\kappa$, a vector space over this countable field has cardinality $\kappa$ exactly when its dimension is $\kappa$, since in general a nonzero space has size $\max (|K|,\dim V)$ when that maximum is infinite. Basis bijections then give categoricity in every uncountable cardinal. Quantifier elimination also shows that these different positive dimensions are elementarily equivalent; independence of an arbitrarily long tuple is not being silently treated as definable over this infinite field.

The distinguished chain generated by zero is countable, and every other component in the classification of models of the successor theory is a countable copy of the integer successor chain. A model of cardinality ${\aleph }_1$ therefore has exactly ${\aleph }_1$ additional components: a finite or countable collection would be countable, while the union of $\kappa$ countable components has size $\max (\kappa ,{\aleph }_0)$.

Given two such models, match their zero chains in the forced way and choose a bijection between their additional components. On each component choose an origin and match corresponding integer iterates; this commutes with successor. The union of these maps is an isomorphism of structures. Existence is given by a disjoint union with ${\aleph }_1$ integer chains. HenceThe same argument works at every uncountable cardinal. Countable models can still differ by the number of extra components, as part (ii) shows.

A $C^1$ diffeomorphism $F:U\to V$ is a bijection for which both $F$ and $F^{-1}$ are continuously differentiable. If a higher regularity class is specified, both maps must belong to that class. In particular, a diffeomorphism is a homeomorphism, and the chain rule gives $D(F^{-1})(F(x))=(DF(x){)}^{-1}$.

A $C^1$ diffeomorphism $F:U\to V$ is a bijection for which both $F$ and $F^{-1}$ are continuously differentiable. If a higher regularity class is specified, both maps must belong to that class. In particular, a diffeomorphism is a homeomorphism, and the chain rule gives $D(F^{-1})(F(x))=(DF(x){)}^{-1}$.

Define $\tilde{f}:X/\mathcal{F}\to Y$ by $\tilde{f}([x])=f(x)$. Its definition is independent of the representative because the classes are precisely the fibres of $f$. It is onto since $f$ is onto, and injective because two points with equal $f$-values lie in the same fibre.

The relation $\tilde{f}\circ p=f$ and part (iii) imply continuity of $\tilde{f}$. Since $X$ is compact and $p$ is continuous, $X/\mathcal{F}$ is compact. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Therefore

A $C^1$ diffeomorphism between open subsets of ${\mathbb{R}}^n$ is a bijection $f:U\to V$ for which both $f$ and $f^{-1}$ are continuously differentiable. More generally, a $C^k$ diffeomorphism requires that regularity for both maps; a smooth diffeomorphism has derivatives of every order. In the $C^1$ setting used here,

Two group actions on $S_1,S_2$ are equivalent if there is a bijection $\varphi :S_1\to S_2$ such that $\varphi (gs)=g\varphi (s)$ for all $g,s$. Such a bijection is equivariant.

The orbit-stabilizer theorem in this form says that the action on $O_s$ is equivalent to the left coset action on $G/G_s$, by $g{G}_s\mapsto gs$. This map is well-defined and injective because $gs=hs$ exactly when $h^{-1}g\in G_s$, and it is onto by the definition of the orbit. It is equivariant by multiplication.

Thus every transitive action is a coset action. Two such actions $G/H$ and $G/K$ are equivalent exactly when $H,K$ are conjugate: stabilizers at different points of one orbit are conjugate, and a conjugacy explicitly supplies a coset-action equivalence. This last classification is what makes counting by orbit types possible.

Two group actions on $S_1,S_2$ are equivalent if there is a bijection $\varphi :S_1\to S_2$ such that $\varphi (gs)=g\varphi (s)$ for all $g,s$. Such a bijection is equivariant.

The orbit-stabilizer theorem in this form says that the action on $O_s$ is equivalent to the left coset action on $G/G_s$, by $g{G}_s\mapsto gs$. This map is well-defined and injective because $gs=hs$ exactly when $h^{-1}g\in G_s$, and it is onto by the definition of the orbit. It is equivariant by multiplication.

Thus every transitive action is a coset action. Two such actions $G/H$ and $G/K$ are equivalent exactly when $H,K$ are conjugate: stabilizers at different points of one orbit are conjugate, and a conjugacy explicitly supplies a coset-action equivalence. This last classification is what makes counting by orbit types possible.

A coordinate transformation is a continuously differentiable bijection from the open set $U$ onto an open subset $T(U)\subset {\mathbb{R}}^2$, with continuously differentiable inverse. Equivalently, it is a $C^1$ diffeomorphism; its Jacobian determinant never vanishes. The inverse function theorem supplies the local inverse, while global injectivity is an additional requirement.