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A finite group is a group whose underlying set has finitely many elements. Its order of a finite group is the number of elements, counting the identity element. The group operation still has to satisfy all the group axioms: merely having a finite list of states does not make a group.

For example, the six congruence classes modulo six form a finite cyclic group under addition, whereas the permutations of three objects form the symmetric group $S_3$, also with six elements. The first is an abelian group; the second is a non-abelian group. Thus the order of a finite group is a useful first measurement, but the composition rule contains information that the count alone loses.

Finiteness makes group structure accessible to exact calculation. A Cayley table for a finite group of order $n$ has $n^2$ entries, and every group element eventually returns to the identity element when repeated. Lagrange's theorem strengthens this observation: the order of a group element, being the size of its cyclic subgroup, divides $n$. The cyclic group of order four and Klein four-group give a small example of how that constraint can distinguish different composition rules.

If every nonidentity element of a finite group has order equal to a prime $p$, its order is a power of $p$. For any prime divisor $q$ of the group order, Cauchy theorem for groups gives an element of order $q$, forcing $q=p$. This statement does not imply that the group is abelian; odd-characteristic upper unitriangular groups provide counterexamples.

If a finite group $G$ is generated by $h$ elements, send the $a$-generators of the genus-$h$ surface group to those generators and its $b$-generators to the identity. The surface relation is satisfied, so this is a surjective group homomorphism. Its kernel defines a connected regular covering space, on which the deck transformation group $G$ acts freely. The lifted orientation and multiplicativity of the Euler characteristic give genus $g=1+|G|(h-1)$.

A finite group of planar rotations about one point is a cyclic group. For a nontrivial group, choose its smallest positive rotation angle $\alpha$. Subtracting multiples of $\alpha$ from any other group angle leaves a group angle in $[0,\alpha )$, hence zero. Applying the same argument to $2\pi$ shows that $\alpha =2\pi /m$ for a positive integer $m$. The identity-only group is cyclic as well.

A subgroup is finitely generated when finitely many of its elements generate it as a group. Even inside a finitely generated group, subgroups need not be finitely generated. Finite generation of a witness subgroup, rather than of the original intersection, is what makes centralizing HNN extensions usable for a benign subgroup.

A finitely presented group is a group admitting a finite group presentation. Finite presentability concerns the existence of such a presentation, not the decidability of its word problem for a group.

A finite group of orientation-preserving isometries of the Euclidean plane or hyperbolic plane is a cyclic group. Take one orbit and its unique minimum-radius enclosing disc. Every group element preserves the orbit and therefore the disc's centre. The orientation-preserving stabilizer of a point acts as planar rotations, so the group embeds in the circle group. In a nontrivial finite subgroup, the least positive rotation angle generates all angles by division with remainder, proving cyclicity. Orientation reversal changes the conclusion: reflection groups need not be cyclic.

A finite quotient is the image of a group under a surjective group homomorphism to a finite group. A group has no nontrivial finite quotients exactly when every homomorphism from it to a finite group is trivial. This property does not prohibit finite or finite-quotient-bearing subgroups.

The finite residual consists of all elements killed by every homomorphism to a finite group. A group is a residually finite group precisely when its finite residual is trivial. Intersecting all finite-index subgroups gives the same result: the normal core of each such subgroup is normal of finite index and lies inside it.

For a group homomorphism $\phi :G\to H$, the group isomorphism $g\ker \phi \mapsto \phi (g)$ givesThe group kernel records which inputs become indistinguishable, and the image records which outputs actually occur. The first isomorphism theorem says that identifying precisely those indistinguishable inputs gives a group isomorphic to the image. It is one of the general isomorphism theorems. ^[ref].

For example, reduction modulo three gives a surjective group homomorphism from the additive cyclic group $\mathbb{Z}/6\mathbb{Z}$ to $\mathbb{Z}/3\mathbb{Z}$. Its group kernel is $\{0,3\}$. The three cosets $\{0,3\}$, $\{1,4\}$, and $\{2,5\}$ become the three outputs zero, one, and two, respectively. ThusThe quotient group diagram shows both the identification and the resulting operation. More generally, fibres of a group homomorphism are cosets of its kernel explains why this map is well-defined and injective; it is surjective onto the image by definition.

A finite group of Euclidean isometries has a common fixed point: average any orbit. Each isometry is an affine map and therefore preserves this finite affine average; applying a group element merely permutes the orbit terms. To see that an isometry is affine, subtract its value at zero and use the polarization identity to show that the resulting map preserves inner products. Images of a standard orthonormal basis then determine a linear orthogonal map.

Foundations of mathematics asks what our objects mean, what we assume about them, and why an argument establishes its conclusion. Set theory gives a common language for collections and functions; mathematical logic makes the structure of statements and deductions precise. A mathematical definition fixes a meaning, an axiom supplies an assumption, and a mathematical proof explains what follows. Separating these roles makes results reusable: once the group axioms are enough for a conclusion, that conclusion applies to every group, regardless of how its members are represented.

The assumptions can change the answer to a familiar calculation. Over the real numbers, $x^2=x$ factors as $x(x-1)=0$, so $x$ is zero or one. The same reasoning holds in any integral domain, because a product can be zero only when a factor is zero. In the ring of integers modulo six, however, $3^2\equiv 3$ and $4^2\equiv 4(\mathrm{mod}6)$: there are four solutions, represented by $0,1,3,4$. The extra solutions are idempotents, and the failed step involves zero divisors. Writing the domain is part of explaining the answer.

This precision also reveals surprising possibilities. A two-dimensional grid of natural numbers is a countable set, while all infinite binary sequences form an uncountable set. The Cantor pairing function supplies a complete enumeration of the first; Cantor's diagonal argument defeats every proposed enumeration of the second. Both arguments are short, yet distinguish two sizes of infinity. The subject's appeal is this ability to turn a vague question into one whose assumptions, construction, and limits can all be inspected.

Every finitely presented group is the fundamental group of a closed connected smooth oriented four-manifold. Build a compact four-dimensional handle decomposition with one zero-handle, a one-handle for each generator of a group and a two-handle for each relator. Turning the handle decomposition upside down shows that its boundary surjects on the fundamental group. Its double of a manifold has the same fundamental group by the Seifert-van Kampen theorem.

A group action is free when only the identity fixes any given point, or equivalently every stabilizer subgroup is trivial. The trivial group acts freely on every space. A nontrivial group acting freely must act by a fixed-point-free map for each nonidentity element; the fixed-point property of even-dimensional complex projective space obstructs such actions. For a finite group acting freely on a Hausdorff manifold, the quotient map is a covering map.

A subset $X$ is a free basis of a group $G$ when the natural map from the free group on $X$ to $G$ is an isomorphism. Equivalently, $X$ generates $G$ and no nonempty freely reduced word in $X\cup X^{-1}$ equals the identity. A generating set is not automatically a free basis.

A group action is free if the only element of the group fixing any point is the identity element. Thus every stabilizer subgroup is trivial. This is a property of an action of any group, not an assertion that the acting group is a free group.

Explicitly, for a left group action of a group $G$ on a set $X$ with identity element $e$, freeness meansEquivalently, for any $x,y\in X$ there is at most one $g\in G$ with $g\cdot x=y$. Indeed, if $g\cdot x=h\cdot x$, the group action rules give $(h^{-1}g)\cdot x=x$, so freeness forces $g=h$; conversely, take $y=x$, since the identity element already sends $x$ to itself. Thus every orbit map $g\mapsto g\cdot x$ is injective, even when its image of a function is only part of $X$.

For a permutation example, let the cyclic group $C_2=\{e,a\}$ act on the set $X=\{1,2,3,4\}$ by $a=(12)(34)$. The nonidentity element exchanges both pairs and fixes no point, so every stabilizer subgroup is $\{e\}$. Its two group orbits are $\{1,2\}$ and $\{3,4\}$. This is a free group action but not a transitive group action: there is no group element sending $1$ to $3$. Each restriction to one group orbit is a simply transitive group action.

The identity element fixes every point in every group action; freeness excludes fixed points only for nonidentity elements. This also differs from a fixed point of a group action, which must be fixed by every group element. In the left diagram there are no such fixed points of a group action. In the right diagram, the same abstract cyclic group $C_2$ acts by $e\cdot x=a\cdot x=x$: all four points are fixed points of a group action, every stabilizer subgroup is $C_2$, and each group orbit is a singleton set. Consequently this second group action is neither free nor transitive. A group action with no fixed points of a group action can still fail freeness: the natural group action of the symmetric group $S_3$ has no point fixed by all of $S_3$, but $(23)$ fixes $1$.

For a finite group, the orbit-stabilizer theorem makes each group orbit of a free group action have cardinality $|G|$. If $X$ is also finite, $|X|$ is therefore a multiple of the group order, with one copy of $|G|$ for each group orbit. The trivial group acts freely on every set, and acts transitively exactly on a singleton set. On the empty set freeness is vacuous; our convention under transitive group action requires a nonempty set for transitivity.