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A nontrivial finite group whose only normal subgroups are the trivial group and the whole group. A cyclic group of prime number order gives an abelian example; $A_5$ gives a nonabelian alternating group example. Finite versions of classical groups can produce further examples after taking suitable subgroups and quotients; small parameters require separate checks.

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.

The fundamental group ${\pi }_1(X,x_0)$ consists of based loops modulo based homotopy. Multiplication follows one based loop and then the other; the constant based loop represents the identity, and path reversal gives the inverse. Passing to homotopy classes makes this operation associative, as explained by well-defined loop concatenation.

Identify the circle with $S^1=\{z\in \mathbb{C}:|z|=1\}$ and choose basepoint $1$. Its based loopshave winding number $n$: positive $n$ counts counterclockwise turns, negative $n$ counts clockwise turns, and $n=0$ gives the constant based loop. Every based loop on the circle is joined by a based homotopy to exactly one ${\gamma }_n$. ConsequentlyThis is Hatcher's Theorem 1.7 in Algebraic Topology, §1.1, pp. 29–30. For example, two counterclockwise turns followed by one clockwise turn represent $2+(-1)=1$, rather than three turns. The winding graph in the figure above makes the signed arithmetic visible.

A proof outline uses the covering map $q:\mathbb{R}\to S^1$, $q(u)=e^{2\pi iu}$. By the path lifting theorem, a based loop has a unique lift starting at zero and ending at an integer $n$. Interpolating its lift to $s\mapsto ns$ produces a based homotopy to ${\gamma }_n$. The homotopy lifting theorem for a covering map keeps that endpoint unchanged during a based homotopy, proving uniqueness.

In the filled disk, every based loop $\gamma$ at $x_0$ has a based homotopy to the constant based loop:Both endpoints remain $x_0$, and the convex set property of the disk keeps every intermediate point inside. Thus ${\pi }_1(D^2,x_0)$ is the trivial group. Together with the deformation retraction above, this computes three concrete answers: the filled disk has trivial fundamental group, while the circle and the annulus have infinite cyclic groups.

Maps have computable algebraic shadows too. The continuous map $p_n:S^1\to S^1$, $p_n(z)=z^n$, fixes $1$ and sends ${\gamma }_k$ to ${\gamma }_{nk}$. Its induced group homomorphism is therefore $k\mapsto nk$, and its mapping degree is $n$. For $n=2$ the image consists of even integers, so this map cannot be a homotopy equivalence. Composing $p_2$ with $p_{-3}$ gives $p_{-6}$: map composition multiplies degrees, whereas concatenating based loops adds winding numbers. Distinct degrees distinguish these maps up to homotopy; this calculation does not classify arbitrary topological spaces.

Historically, Poincaré's 1895 Analysis Situs introduced the fundamental group; its original §12 explicitly names it. The CNRS transcription of Poincaré's §12, Groupe fondamental provides a direct view of this early formulation.

Up to group isomorphism, the finite groups with at most six elements are the trivial group, the cyclic groups $C_2,C_3,C_4,C_5,C_6$, the Klein four-group $C_2\times C_2$, and the symmetric group $S_3$.

A group of prime cardinality is a cyclic group, since Lagrange's theorem forces any nonidentity element to generate it. For four elements, either an element has order of a group element four, or every nonidentity element is an involution. In the latter case the group of exponent two is abelian identity shows that two distinct nonidentity elements $a,b$ generate $\{e,a,b,ab\}\cong C_2\times C_2$.

For a group of order six, either an element generates it or some element $a$ has order of a group element three: otherwise all nonidentity elements would be involutions, producing a four-element subgroup in contradiction with Lagrange's theorem. Pairing each element with its inverse element gives an involution $b$ outside $⟨a⟩$. The subgroup $⟨a⟩$ has index of a subgroup two and is a normal subgroup, so $ba{b}^{-1}=a$ or $a^{-1}$. The commuting case makes $ab$ have order of a group element six, whereas the second case gives the multiplication rules of $S_3$ on the six distinct elements $a^i,a^{i}b$, $0\le i<3$.

All these groups embed in $S_4$ except $C_5$ and $C_6$. The former is excluded by Lagrange's theorem; the latter is excluded because the disjoint cycle decomposition of a permutation of four letters cannot have least common multiple of cycle lengths equal to six.

A free product $A\ast B$ is nontrivial as a splitting when both factors differ from the trivial group. Alternating words $(ab{)}^r$ with $a,b\ne 1$ in different factors show that every such splitting gives an infinite group. The distinct normal forms $ab$ and $ba$ also show that the group is not an abelian group.

The order of a finite group is its cardinality, written $|G|$. The identity element is included: $|\mathbb{Z}/6\mathbb{Z}|=6$, and the trivial group has order one, rather than zero.

Keep this count distinct from the order of a group element. In the additive finite cyclic group $\mathbb{Z}/6\mathbb{Z}$, the element $1$ has order six, $2$ has order three, $3$ has order two, and $0$ has order one. Each element's order is the least positive number of repetitions needed to return to the identity element $0$. The order of an element of a finite cyclic group computes these periods directly.

The count also constrains subgroups: Lagrange's theorem gives $|G|=[G:H]|H|$ for a subgroup $H$ of a finite group $G$, where the index of a subgroup counts its cosets. For $H=\{0,2,4\}$ inside $\mathbb{Z}/6\mathbb{Z}$, the two cosets are $\{0,2,4\}$ and $\{1,3,5\}$, so $6=2\cdot 3$. Divisibility is a necessary constraint; it does not imply that every divisor occurs as the order of a group element, as the Klein four-group shows.

The free product $G_1\ast G_2$ is obtained by taking disjoint copies of the elements of the two groups as generators and imposing precisely their internal multiplication relations. Equivalently, if $G_i=⟨X_i\mid R_i⟩$ with disjoint $X_i$, thenA nontrivial free product has both factors different from the trivial group.

The normal form theorem for a free product says that every element has a unique expressionwhere the empty expression represents the identity. The $g_j$ are called syllables in a free product. Thus nonempty alternating products cannot be the identity, and the natural maps of the factors into the free product are injective.

For existence, multiply adjacent syllables from the same factor and delete identity syllables until the word alternates. Each change uses a defining relation and decreases the number of syllables. To establish uniqueness without assuming that these reductions are confluent, let $\mathcal{W}$ be the set of alternating nonidentity syllable sequences, including the empty sequence. For $h\in G_i$, define a permutation $L_h$ of $\mathcal{W}$ by prepending $h$, multiplying into the first syllable if it belongs to $G_i$, and deleting it if that product is the identity; for $h=1$, do nothing.

The multiplication law inside $G_i$ gives $L_h{L}_k=L_{hk}$ for $h,k\in G_i$, and $L_h^{-1}=L_{h^{-1}}$. This can be checked at the first syllable: if a product deletes it, the next syllable belongs to the other factor and is treated as a new first syllable. Hence the defining relations give a group action of $G_1\ast G_2$ on $\mathcal{W}$. An alternating word $g_1\cdots g_r$ sends the empty sequence to its own syllable sequence. Two such words representing the same group element induce the same permutation, and therefore have identical sequences. This proves the normal form theorem for a free product.

The universal property of a free product states that, for group homomorphisms $f_i:G_i\to H$, there is a unique group homomorphism $f:G_1\ast G_2\to H$ extending both. Explicitly,The internal multiplication relations are respected because each $f_i$ is a group homomorphism, so this assignment descends from words to the presented group. Alternatively, multiplication of two normal forms is concatenation followed by precisely those internal relations, which do not change its value in $H$. Uniqueness holds because the two factors generate the free product.

Part (c) shows that every finitely presented group embeds in a finitely presented group with no nontrivial finite quotients of a group. Consequently this property also has no possible forbidden witness $A_{-}$. ThereforeBoth properties have positive witnesses, for example $C_2$ and the trivial group respectively. It is the required embedding obstruction that fails in each case.

The complete classification up to group isomorphism isHere $C_1$ is the trivial group. The prime number cases follow from i. For four elements, an element of order of a group element four generates $C_4$. If there is none, Lagrange's theorem makes every nonidentity element have order of a group element two. Then iv produces a four-element subgroup, which must be all of $G$, giving the Klein four-group.

For six elements, an element of order of a group element six gives $C_6$. Suppose there is none. If there were no element of order of a group element three either, all nonidentity elements would have order of a group element two. Part iv would then produce a subgroup of cardinality four, contradicting Lagrange's theorem. Thus choose $a$ of order of a group element three. There is also an involution $b$: otherwise the five nonidentity elements could be paired into distinct pairs $\{g,g^{-1}\}$, impossible for an odd number of elements.

The subgroup $H=⟨a⟩$ has index of a subgroup two, so it is a normal subgroup. Indeed, for $g\notin H$, both $gH$ and $Hg$ are the complement of $H$, and for $g\in H$ both equal $H$. Consequently $ba{b}^{-1}$ is either $a$ or $a^{-1}$. The first possibility makes $a,b$ commute, and $ab$ then has order of a group element six: $(ab{)}^k=e$ forces $a^k=b^{-k}\in ⟨a⟩\cap ⟨b⟩=\{e\}$, so both $3$ and $2$ divide $k$. This contradicts our assumption. ThereforeEvery element is uniquely $a^i$ or $a^{i}b$, with $0\le i<3$, and the relation $ba=a^{-1}b$ determines all their products. Sending $a$ to $(123)$ and $b$ to $(12)$ gives a group isomorphism with the symmetric group $S_3$. This completes the classification of groups of order at most six.

Of these groups, exactly $C_1,C_2,C_3,C_4,C_2\times C_2,S_3$ embed in $S_4$. Concrete subgroups areThere is no $C_5$ subgroup, since $5$ does not divide $24$. There is no $C_6$ subgroup either: the order of a group element in a symmetric group is the least common multiple of the lengths in its disjoint cycle decomposition, and on four letters the only possibilities are $1,2,3,4$.

Label the vertices of a regular tetrahedron $1,2,3,4$, and include all its Euclidean isometries, both rotations and reflections. The following list describes every subgroup of cardinality at most six, allowing arbitrary relabelling of the vertices. The numbers count actual subgroups, rather than just their group isomorphism types.

Here is why this is exhaustive. Part b leaves only the trivial group, $C_2,C_3,C_4,C_2\times C_2,S_3$. The disjoint cycle decomposition gives exactly the two listed kinds of involutions, the three-cycles, and the four-cycles, accounting for all the cyclic subgroups. In a Klein four-group, the nonidentity elements commute. If it contains $(12)$, any other nonidentity element must be a commuting involution, hence $(34)$ or $(12)(34)$; either choice yields the displayed opposite-edge subgroup. If it has no transposition, its three nonidentity elements must be all three double transpositions, yielding the rotational subgroup.

Finally, an $S_3$ subgroup has a unique cyclic subgroup of cardinality three. Its two three-cycles fix the same unique vertex. Every element of the $S_3$ subgroup preserves this vertex, since conjugation preserves that unique cyclic subgroup. Thus it is the full stabilizer subgroup of that vertex. The counts in the list are consequently exhaustive, givingThe two kinds of $C_2$ and the two kinds of Klein four-groups have different geometric behavior despite having the same abstract group isomorphism type within each pair. This is the distinction recorded by the small subgroups of tetrahedral symmetry.

If $G=\{e\}$, choose ordinary addition on $\mathbb{Z}$, giving the additive group of integers. Every power function is the unique identity function on the singleton $G$, so the group action axioms hold:Thus ordinary addition supplies the required group operation on $\mathbb{Z}$. Together with ii, existence is equivalent to $G$ being the trivial group.

For distinct $i,j\ne 1$, direct conjugation givesThus the given star transpositions generate every transposition. Every cycle can in turn be expressed asBy the disjoint cycle decomposition, every permutation is a product of cycles, hence of transpositions, hence of the given star transpositions. ThereforeFor $n=1$, the empty generating set of a group generates the trivial group $S_1$.

Yes. Take $X=S^1$, the circle, and $A=\{1\}\subset S^1$. The constant function $r(z)=1$ is a continuous map and satisfies $ri={\mathrm{id}}_A$, so $A$ is a retract of $X$.

However, their fundamental groups areA homotopy equivalence between path-connected spaces induces a group isomorphism of their fundamental groups, allowing the usual change of base point. The infinite cyclic group is not the trivial group. Therefore a retract need not have the homotopy type of its ambient space.

Suppose a binary operation on the underlying set $\mathbb{Z}$ makes it a group, and $\rho (n,x)=x^n$ is a group action on another group $G$. Every acting element must induce a bijection, with inverse function given by its acting inverse element. The integer $0$ instead induces the constant function $x\mapsto e$, where $e$ is the identity element of $G$, so $G$ must be the trivial group. Conversely, on the trivial group, ordinary addition on $\mathbb{Z}$ supplies such a group action.

A subgroup $H\le G$ is a subset of a group $G$ that is itself a group under the same group operation. It must contain the identity element of $G$ and be closed under products and inverse elements. Associativity is inherited from $G$, so it needs no new proof. The whole group and the trivial group $\{e\}$ are always subgroups. A proper subgroup satisfies $H\ne G$.

A useful single-condition test isTo see why it works, choose $a\in H$; then $a{a}^{-1}=e\in H$. Taking $a=e$ gives every inverse element $b^{-1}\in H$. Applying the test to $a$ and $b^{-1}$ then gives $ab\in H$. Conversely, every subgroup satisfies the test by closure under products and inverse elements. This also makes the additive version memorable: a nonempty subset of an additive group is a subgroup exactly when it is closed under differences. See McKernan's MIT lecture 2, Proposition 2.3, for the equivalent closure criterion.

For the square dihedral group $D_4$ above, $R=\{e,r,r^2,r^3\}$ is a cyclic subgroup: products add exponents modulo four, and $(r^k{)}^{-1}=r^{-k}$. The smaller collections $\{e,r^2\}$ and $\{e,s\}$ are subgroups of group order two. But $\{e,r\}$ is not a subgroup, since $r\cdot r=r^2$ lies outside it. A plausible-looking list of symmetries therefore need not constitute a complete collection of reversible operations.

Inside $(\mathbb{Z},+)$, the even numbers $2\mathbb{Z}$ form a subgroup: $2m-2n=2(m-n)$ is an even number. Its two cosets consist of the even numbers and the odd numbers, so its index of a subgroup is two. The parity group homomorphism $k\mapsto k\mathrm{mod}2$ has $2\mathbb{Z}$ as its kernel of a group homomorphism. A subgroup can thus identify exactly which information a composition-preserving map forgets.

Subgroups also describe residual symmetry. In the group action of $D_4$ on the four vertices, the stabilizer subgroup of $(1,1)$ is $\{e,sr\}$: the diagonal reflection $sr$ fixes that vertex, and the other six motions move it. Requiring a feature to stay fixed still leaves a group, because both composition and reversal preserve that requirement. This is why subgroups are useful for studying a group action one feature at a time.