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Lagrange's theorem states that for a subgroup $K$ of a finite group $G$,in particular, $|K|$ divides $|G|$. Here $[G:K]$ is the subgroup index, the number of left cosets of $K$.

The order of a group element $g\in G$ is the least positive integer $k$ such that $g^k=e$. It exists because $G$ is finite. The cyclic subgroup $⟨g⟩$ consists of exactly $e,g,\dots ,g^{k-1}$, so has order $k$. Applying Lagrange's theorem to this subgroup gives

Let the finite set $G/H$ contain $r$ distinct left cosets. By i, they form a disjoint partition of a set $G$, and by ii each has $|H|$ elements. If $H$ is also finite, counting this finite disjoint set union givesThus $G$ is a finite group. This is the counting argument behind Lagrange's theorem.

A conjugacy class $C(x)$ is a singleton exactly when $gx{g}^{-1}=x$ for every $g\in G$, equivalently when $gx=xg$ for every $g$. Thus the singleton conjugacy classes correspond exactly to elements of the center of a group $A$, and there are $|A|$ of them.

Every other conjugacy class has at least two elements. Since the conjugacy classes form a partition of a set $G$, if $r$ is the number of these remaining classes then $2r\le |G|-|A|$. ConsequentlyThis conjugacy class count bound from the center needs no classification of the finite group.

The Klein four-group has multiplication tableSince it is abelian, every conjugacy class is a singleton: $\{1\},\{a\},\{b\},\{c\}$. An irreducible representation is a nonzero representation whose only invariant subspaces are zero and the whole representation space. For a finite group, the number of inequivalent complex irreducible representations equals the number of conjugacy classes, so there are four here.

A one-dimensional representation sends the involutions $a,b$ to signs, independently. All four choices areThey are distinct irreducible representations, exhausting the four possibilities. The character table of the Klein four-group isThe specified diagonal matrices obey $D(a{)}^2=D(b{)}^2=I$, commute, and satisfy $D(a)D(b)=D(c)$, so the group relations prove that $D$ is a group representation. Its coordinate axes are invariant lines carrying respectively ${\chi }_{01},{\chi }_{10},{\chi }_{11}$. ThereforeThe trivial representation ${\chi }_{00}$ is absent. If $H$ denotes the four-row character matrix above, direct row products give $H{H}^T=4I$. This verifies character orthogonality:The character of $D$ is $(3,-1,-1,-1)$, whose inner products with the four rows are $0,1,1,1$, independently confirming the decomposition. All these irreducibles are real as well: commuting real involutions can be simultaneously diagonalized into their real sign eigenspaces.

For a subgroup $H$ of a finite group $G$, distinct left cosets are disjoint and partition $G$. Each coset $gH$ has exactly $|H|$ elements because $h\mapsto gh$ is a bijection. Counting the partition proves Lagrange's theorem:A cyclic group is generated by a single element. In a group of order four, an element of order four generates the whole group, giving $C_4$. Otherwise all three nonidentity elements have order two. Choose two of them $a,b$; cancellation makes $ab=c$ the third. Since $c^{-1}=c$, $(ab{)}^{-1}=b^{-1}{a}^{-1}=ba=ab$. The remaining products then give the Klein four-group. This proves that there are precisely two groups of order four up to isomorphism, distinguished by whether an element of order four exists.

For the square, take $r$ to be counterclockwise rotation by $90^{\circ }$ and $s$ reflection in the horizontal axis. They satisfy $r^4=s^2=e$ and $sr=r^{-1}s$. The dihedral group here is $D_4$ of order eight, with elements $e,r,r^2,r^3,s,rs,r^{2}s,r^{3}s$. Both generators are necessary: one generator would give a cyclic, hence abelian, group, while $r$ and $s$ do not commute. In the following Cayley table, the row acts after the column; the formula $(r^a{s}^b)(r^c{s}^d)=r^{a+(-1{)}^{b}c}{s}^{b+d}$, with exponents reduced modulo four and two, computes every entry.The largest cyclic subgroup is $⟨r⟩=\{e,r,r^2,r^3\}$, of order four. Every reflection has order two and there is no element of order eight. The other subgroups of order four areBoth are Klein four-groups, generated by the commuting involutions $r^2$ and one reflection. For completeness, any order-four subgroup containing $r$ or $r^3$ equals $⟨r⟩$. Otherwise it contains a reflection $f$; the product of two distinct reflections is a nonidentity rotation, which must be $r^2$ if odd powers of $r$ are excluded. This forces the subgroup to be $\{e,r^2,f,r^{2}f\}$, giving exactly the two displayed choices. It is the square case of the classification of subgroups of a dihedral group.

Write $A(a,b,c)$ with $a$ in entry $(1,2)$, $b$ in entry $(1,3)$ and $c$ in entry $(2,3)$, and perform all arithmetic in the finite field ${\mathbb{F}}_5$. Direct multiplication givesThus the set is closed. Associativity comes from matrix multiplication. The identity element is $A(0,0,0)$, and solving the product formula for it givesBoth orders of multiplication with this inverse give the identity. Hence these matrices form a finite group, an instance of the Heisenberg group over a prime field. Each coordinate has five independent possibilities, with distinct triples giving distinct matrices, soThe explicit coordinate convention matters because the central coordinate is $b$ here; other descriptions of the Heisenberg group over a prime field may place the central coordinate last.

For a finite group over the complex numbers, the sum of the squared dimensions of its inequivalent irreducible representations equals the group order. There is always a one-dimensional trivial representation, and part (b) supplies an irreducible representation of dimension three. Thus the remaining squared dimensions sum toThey must be two further dimensions equal to one. HenceThe number of inequivalent irreducible representations equals the number of conjugacy classes, so $A_4$ has four classes. There are only three cycle types in part (a); consequently cycle type cannot classify conjugacy within $A_4$.

More explicitly, the identity is a singleton class, and conjugation by three-cycles permutes the three double transpositions transitively. For a three-cycle, the centralizer in $A_4$ consists of its three powers, so its conjugacy class has $12/3=4$ elements. The eight three-cycles therefore split into two classes of four:For example, $(123)$ and $(132)$ have the same cycle type but are not conjugate in $A_4$. Any permutation conjugating the first to its inverse must fix its unique fixed vertex and reverse the orientation of the remaining three labels, so it is odd. They are conjugate in $S_4$, illustrating exactly why restricting the ambient group changes the answer.

The group order is its cardinality, $\boxed{|G|=\#\{g:g\in G\}}$. The order of a group element $g$ iswhere $e$ is the identity element. In a finite group such an $n$ always exists, since two nonnegative powers of $g$ must coincide.

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 finite group is solvable if its derived series, defined by $G^{(0)}=G$ and $G^{(j+1)}=[G^{(j)},G^{(j)}]$, reaches the identity after finitely many steps. Equivalently it has a subnormal series with abelian factor groups. The factors need not all have prime order for this definition.

For a group action, the orbit and stabilizer areThe orbit-stabilizer theorem for a finite group states

A composition series is a finite chainwhose inclusions are strict and whose quotients $G_i/G_{i-1}$ are simple. These quotients are the composition factors. The normal subgroup condition is relative to the next member, rather than necessarily to all of $G$.

Induct on the order of a finite group. The trivial group has the empty composition series. Otherwise choose a maximal normal subgroup $N⪇G$, which exists because there are finitely many subgroups. The subgroup correspondence theorem identifies normal subgroups of $G/N$ with normal subgroups of $G$ containing $N$, so maximality makes $G/N$ a simple group. By induction $N$ has a composition series; appending $G$ gives one for $G$.

For example, the nonabelian symmetric group $S_3$ haswith composition factors $C_3$ and $C_2$, both abelian.

Supposeare two factorizations into primes. Repeatedly apply the cyclic-subgroup construction of part (c), first removing $p_r$, then $p_{r-1}$, and so on. This constructs a composition series for $C_n$ whose composition factors are $C_{p_1},\dots ,C_{p_r}$. The second factorization constructs another composition series with composition factors $C_{q_1},\dots ,C_{q_s}$.

By the Jordan–Hölder theorem, $r=s$ and these composition factors agree up to group isomorphism and permutation. Since isomorphic finite groups have the same order, the corresponding primes agree. This proves uniqueness of prime factorization up to permutation, using existence of prime factorizations but not their uniqueness.

For a finite group action, Burnside's lemma givesPair each vertex with its opposite face. This correspondence is equivariant under every tetrahedral symmetry. Each pair has six possible combined color states: three face colors and two vertex colors. Thus we are coloring four positions with six symbols; the two kinds of coloring must be counted together.

The group of rotations is the alternating group $A_4$. The identity has four cycles and fixes $6^4$ colorings. Its three half-turns have cycle type $(2,2)$ and each fix $6^2$. Its eight third-turns have cycle type $(3,1)$ and each also fix $6^2$. Consequently the rotation count isAllowing mirror equivalence gives the full tetrahedral symmetry group $S_4$. Now all permutations of the four paired positions are allowed, so a coloring is determined by a multiset of four symbols chosen from six. stars and bars givesFor an independent Burnside check, the five cycle types of $S_4$ give numerator $6^4+6\cdot 6^3+3\cdot 6^2+8\cdot 6^2+6\cdot 6=3024$, whose quotient by 24 is 126. This is paired face and vertex colorings of a tetrahedron.

For a group action, the orbit and stabilizer areThe orbit-stabilizer theorem for a finite group states

With local path connectedness, the intended conclusion holds. Choose a base point $x_0$ and the standard covering mapThe fundamental group of the circle is $\mathbb{Z}$. Since the image of a finite group under a group homomorphism is finite and $\mathbb{Z}$ has no nontrivial finite subgroups,is trivial. If $X$ is also locally path-connected, the lifting criterion for a covering space supplies a continuous map $\tilde{f}:X\to \mathbb{R}$ with $p\tilde{f}=f$, because $p_{\ast }{\pi }_1(\mathbb{R})=0$. The homotopyruns from $f$ to the constant function $f(x_0)$. This proves the claimed null-homotopy under that additional hypothesis, in particular for CW complexes and the topological graphs considered later in the paper.

For arbitrary path-connected spaces, the printed assertion is false. Here is an explicit Warsaw circle counterexample, including the obstruction that the usual lifting argument misses. In the Euclidean plane takeJoin $P=(1,0)$ to $Q=(0,-1)$ by the polygonal arc $A$ passing successively through $(2,0),(2,-2),(0,-2)$, and put $W=G\cup L\cup A$ with its subspace topology. The arc meets $G\cup L$ only at $P,Q$. The topologist's sine curve $G\cup L$ has two path components, but the added arc joins them, making $W$ path-connected.

Every loop in $W$ is contained infor some $\delta >0$. To justify the assertion, a path cannot reach $L$ along the oscillating tail: the intermediate value theorem for its $x$-coordinate would force its $y$-coordinate to attain both $1$ and $-1$ arbitrarily close to the supposed arrival time, contradicting continuity. The only alternative route from the tail to $L$ uses $P$ and the joining arc. If a loop visited tail points with $x\to 0$, a convergent subsequence of their parameter values would approach a time at which the loop lies in $L$. On a sufficiently small parameter interval around that time its image avoids $P$. That would give a forbidden path from $L$ into the tail without passing through $P$. Thus those tail coordinates have a positive lower bound. The space $T_{\delta }$ is a finite tree, in fact a single arc, so every such loop contracts. ConsequentlyNow define a continuous map $h:W\to S^1$ to be $1$ on $G\cup L$, and to make one full turn on $A$: for a parametrization $a:[0,1]\to A$ from $P$ to $Q$, set $h(a(t))=e^{2\pi it}$. The pieces $G\cup L$ and $A$ are closed in $W$, and the formulas agree at $P,Q$, so the pasting lemma for closed subspaces proves continuity.

There is no continuous lift $\tilde{h}:W\to \mathbb{R}$. Any such lift is integer-valued on $G\cup L$, so it is constant on each of the connected spaces $G,L$. Since $G$ accumulates on $L$, continuity forces those two constants to be the same. But along $A$ a lift of $e^{2\pi it}$ has the form $t+k$, giving values at $Q$ and $P$ differing by $1$, a contradiction. Finally, if $h$ had a null-homotopy, the homotopy lifting theorem for a covering map, starting with a lift of the final constant function and reversing time, would produce a continuous lift of $h$. This theorem needs no local path connectedness of the parameter space. Hence $h$ is not null-homotopic, although $W$ has finite, indeed trivial, fundamental group.

The native PDF explicitly assumes only path connectedness here. The missing local path connectedness cannot be inferred from the finiteness of the fundamental group.

The largest normal p-subgroup of a finite group. Products of normal p-subgroups are again normal p-subgroups, since $|AB|=|A||B|/|A\cap B|$. Thus the subgroup generated by all such subgroups is itself a p-subgroup. It is a characteristic subgroup.

A group is poly-(cyclic or finite) when it has a finite subnormal series $1=G_0◃G_1◃\cdots ◃G_r=G$ in which each quotient $G_i/G_{i-1}$ is a cyclic group or a finite group. Each term need only be normal in the next term. The polycyclic group condition instead requires every factor to be cyclic.

For a finite group with trivial centre of a group and a prime divisor $p$ of its order, some nonidentity conjugacy class has size coprime to $p$. Otherwise the class equation would give $|G|\equiv 1(\mathrm{mod}p)$. Primality is essential: in the symmetric group $S_3$, neither nonidentity class size is coprime to the composite divisor $6$.

A finite group is $m$-quasirandom if every nontrivial irreducible representation over the complex numbers has degree at least $m$. The trivial representation is excluded. Large $m$ suppresses correlations of products of arbitrary subsets, through product mixing in a quasirandom group.