Finite group theory

Finite group theory studies groups with finitely many elements.

Upper-unitriangular three-by-three matrices over the prime finite field form the finite analogue of the real Heisenberg group. In coordinates their multiplication is $(a,b,c)(a^{\prime },b^{\prime },c^{\prime })=(a+a^{\prime },b+b^{\prime },c+c^{\prime }+a{b}^{\prime })$. The order is $p^3$, and $(a,b,c{)}^k=(ka,kb,kc+\left(\genfrac{}{}{0ex}{}{k}{2}\right)ab)$. For odd $p$, every nonidentity element has order $p$, but the group is nonabelian. For $p=2$ the exponent statement changes, so odd characteristic is essential for the nonisomorphic Gassmann equivalent regular subgroups construction.

A regular tetrahedron has a proper rotational symmetry group $T$ of order 12, isomorphic to the alternating group $A_4$ acting on its four vertices. Its full spatial symmetry group $T_d$ has order 24 and is isomorphic to the symmetric group $S_4$; the additional operations reverse orientation. Three-fold vertex/face-axis rotations and half-turns about opposite-edge midpoints generate the proper group. The distinction matters when a field symmetry uses only proper rotations, but a scalar density also admits reflections.

Up to group isomorphism, the groups of order eight are $C_8$, $C_4\times C_2$, $C_2^3$, $D_8$ and $Q_8$, with $D_8$ here denoting the eight-element dihedral group. An element of order eight gives the first case. If every nonidentity element has order two the group is an elementary abelian group. Otherwise, after excluding elements of order eight, a cyclic subgroup of order four has index two; an element outside it either commutes with its generator or inverts it. In the latter case its square is either the identity or the central involution, giving the dihedral group or quaternion group.

Every finite group of odd order is a solvable group. In particular, a nonabelian simple group has even order and therefore has an involution by Cauchy's theorem for finite groups. This is a deep theorem, not an elementary consequence of the Sylow theorems.

The alternating conjugacy class splitting criterion gives seven conjugacy classes in $A_6$. Their cycle types and sizes are $1^6:1$, $2^2{1}^2:45$, $31^3:40$, $3^2:40$, $42:90$, and two $51$ classes of size 72. Representatives of the split classes are a 5-cycle and its square. Their sizes sum to 360. A normal subgroup must be a union of these classes containing the identity, so these explicit sizes can prove simplicity using Lagrange's theorem.

For $n\ge 2$ and $1\le k<n$, an $n$-cycle $c=(12\dots n)$ and $t=(11+k)$ generate $S_n$ exactly when $\gcd (n,k)=1$. Their conjugate transpositions connect labels differing by $k$ modulo $n$. This graph is connected exactly at coprime separation, so transpositions on a connected graph generate the symmetric group. Otherwise residue classes modulo $\gcd (n,k)$ form a nontrivial block system preserved by both generators.

Associate a transposition to each edge of a finite connected graph. For a simple path $v_0,\dots ,v_{\ell }$, let $t_j=(v_{j-1}{v}_j)$. Then $t_1\cdots t_{\ell -1}{t}_{\ell }{t}_{\ell -1}\cdots t_1=(v_0{v}_{\ell })$, using rightmost-first composition. Thus edge transpositions generate every transposition, and hence the full symmetric group.

The parity of a permutation records whether it is a product of an even or odd number of transpositions. It is well-defined and equals the sign of the corresponding permutation matrix.

The sign homomorphism sends an even permutation to $1$ and an odd permutation to $-1$; its kernel is the alternating group.

A symmetric group is the group of all bijections from a set to itself, with function composition as its operation.

A permutation cycle $(a_1{a}_2\dots a_k)$ sends $a_i$ to $a_{i+1}$, sends $a_k$ to $a_1$, and fixes every other element. Its order is $k$.

The cycle type of a permutation is the multiset of lengths in its decomposition into disjoint permutation cycles, including one-cycles for fixed points. Two permutations in the same symmetric group are conjugate exactly when they have the same cycle type.

A three-cycle is a permutation cycle of length three. On any chosen set of three letters, the two possible three-cycles are inverses of one another.

Permutation cycles with disjoint supports commute. The order of their product is the least common multiple of their lengths.

The support of a permutation is the set of elements it does not fix. A permutation has finite support when this set is finite.

The sign homomorphism $\mathrm{sgn}:S_n\to \{\pm 1\}$ is the determinant of the corresponding permutation matrix. A transposition has sign $-1$, so this also proves that the parity of any transposition decomposition is well-defined.

An odd permutation has sign $-1$, equivalently it is a product of an odd number of transpositions.

An even permutation has sign $+1$, equivalently it is a product of an even number of transpositions.

The quaternion group $Q_8=\{\pm 1,\pm i,\pm j,\pm k\}$ is a nonabelian Dedekind group.

The least size of a finite set on which the quaternion group acts faithfully is eight. Every nontrivial subgroup of $Q_8$ contains its central involution $-1$. In an action on fewer than eight points, the orbit-stabilizer theorem makes every stabilizer subgroup nontrivial, so $-1$ fixes every point. This rules out a faithful group action, even with several group orbits. The regular action on eight points is faithful by Cayley theorem.

The dicyclic group of order $4n$ has generatorsEvery element has a unique form $a^k$ or $a^{k}b$, with $0\le k<2n$.

Four one-dimensional characters and $n-1$ two-dimensional characters exhaust a dicyclic group of order $4n$. In the two-dimensional rows, the values at $a^r$ are $2\cos (\pi jr/n)$ and those on the two noncyclic cosets are zero. The one-dimensional rows depend on whether $n$ is odd or even.

A finite p-group is a finite group whose order is a power of one prime number $p$.

The nontrivial center of a finite p-group theorem implies that the center of a group of order $p^2$ has order $p$ or $p^2$. If it has order $p$, the quotient by the center has prime order and is cyclic; a cyclic quotient by the center forces the whole group to be abelian, contradicting the assumed size of its center. Thus the group is abelian. An element of order $p^2$ makes it cyclic. Otherwise choose $a\ne 1$ and $b\notin ⟨a⟩$; their order-$p$ cyclic subgroups intersect trivially and generate the direct product of groups $C_p\times C_p$. The two possibilities are nonisomorphic because only the cyclic one has an element of order $p^2$.

Every proper subgroup $H$ of a finite p-group is properly contained in its normalizer. Let $H$ act on the left cosets $G/H$ by left multiplication. The fixed cosets are those represented by elements of $N_G(H)$, so there are $[N_G(H):H]$ of them. All other orbits of a group action have sizes divisible by $p$. Since $[G:H]$ is divisible by $p$ and at least one coset is fixed, the fixed-coset count is a positive multiple of $p$. This also proves $[N_G(H):H]\ge p$.

A nontrivial normal subgroup $N$ of a finite p-group meets the center of a group nontrivially. The conjugation action of $G$ on $N$ has singleton orbits exactly at $N\cap Z(G)$. The orbit-stabilizer theorem makes every other orbit size divisible by $p$, so $|N\cap Z(G)|\equiv |N|\equiv 0(\mathrm{mod}p)$. Since the intersection contains the identity, its size is at least $p$.

Every nontrivial finite p-group has a nontrivial center. The conjugation action partitions the group into conjugacy classes whose noncentral sizes are positive powers of $p$; the class equation then implies $|Z(G)|\equiv |G|\equiv 0(\mathrm{mod}p)$.

The Frattini subgroup of a finite p-group is the intersection of its maximal subgroups. It equals $G^p[G,G]$.

The Frattini quotient of a finite p-group is an elementary abelian p-group. Its dimension over ${\mathbb{F}}_p$ is the minimum number of generators of $G$; in particular, it has dimension at least two when $G$ is noncyclic.

A nontrivial simple group has no normal subgroups other than the identity subgroup and itself.

For a faithful primitive group action, suppose a point stabilizer has an abelian normal subgroup whose conjugates generate the whole group. Every nontrivial normal subgroup then contains the derived subgroup. Indeed the normal subgroup is transitive, and all those conjugates have the same image in the quotient, making the quotient abelian. If the whole group is nontrivial and perfect, it is simple.

A normal subgroup of $A_6$ has order dividing 360 and consists of the identity plus whole conjugacy classes of the alternating group on six letters. If it omits the odd-sized class of size 45, its order is an odd divisor and no nontrivial class sum fits. If it includes that class, the possible class sums not exceeding 180 are $46,86,118,126,136,158,176$, none dividing 360. Every proper subgroup has order at most 180; therefore $A_6$ is a simple group.

The alternating group $A_n$ is simple for every $n\ge 5$. In particular it has no proper subgroup of index two, since every index-two subgroup is normal.

A finite nonabelian simple group is a finite group that is a nonabelian group and a simple group. Every such group is a perfect group because its commutator subgroup is a nontrivial normal subgroup.

The conjugacy classes of $A_5$ have sizesA normal subgroup is a union of these classes containing the identity. No proper nontrivial such union has size dividing $60$, so Lagrange's theorem proves that $A_5$ is simple.

Every finite nonabelian simple group has order at least 60; the alternating group A5 attains this bound.

The group $G{L}_n({\mathbb{F}}_q)$ consists of invertible $n$ by $n$ matrices over the finite field ${\mathbb{F}}_q$.

For prime $p$, every order-$p$ element of $G{L}_2({\mathbb{F}}_p)$ is conjugate to the displayed nonidentity unipotent matrix. Its cyclic subgroup has vector group orbits of sizes one or $p$; counting all $p^2$ vectors implies at least $p$ fixed vectors. Choose a nonzero fixed vector as first basis vector. The matrix becomes triangular with first diagonal entry one; its other diagonal entry satisfies $d^p=1$ and hence $d=1$ in the prime finite field. The off-diagonal entry is nonzero and can be rescaled to one. All matrices involved have invertible changes of basis.

The general linear group over a finite field $G{L}_2({\mathbb{F}}_2)$ acts faithfully on the four vectors of ${\mathbb{F}}_2^2$, since a matrix fixing all vectors fixes the standard basis. The zero vector is fixed by every element; the induced action on the three nonzero vectors identifies this six-element group with $S_3$. Swapping the two coordinates fixes $(0,0)$ and $(1,1)$ and exchanges the other two vectors, so its four-point permutation is odd. This action therefore embeds the group into $S_4$ with nontrivial composite sign homomorphism.

Choosing linearly independent columns successively gives

The upper unitriangular group consists of upper triangular matrices with every diagonal entry equal to one. Over ${\mathbb{F}}_p$ it has order $p^{n(n-1)/2}$ and is a Sylow subgroup of $G{L}_n({\mathbb{F}}_p)$.

For a three-dimensional strictly upper triangular matrix $A$, the binomial theorem stops after $A^2$. Over a finite field of odd prime characteristic $p$, both coefficients $p$ and $\left(\genfrac{}{}{0ex}{}{p}{2}\right)$ vanish, so every nonidentity matrix in $U{T}_3({\mathbb{F}}_p)$ has order $p$. In characteristic two, a matrix with nonzero $A^2$ instead has order four.

The upper unitriangular group $U{T}_3({\mathbb{F}}_3)$ consists of $M(a,b,c)=\left(\begin{array}{ccc}1& a& c\\ 0& 1& b\\ 0& 0& 1\end{array}\right)$ for $a,b,c\in {\mathbb{F}}_3$. Its multiplication is $M(a,b,c)M(a^{\prime },b^{\prime },c^{\prime })=M(a+a^{\prime },b+b^{\prime },c+c^{\prime }+a{b}^{\prime })$. The elements $M(1,0,0)$ and $M(0,1,0)$ do not commute. Yet every element is $I+N$ with $N^3=0$, so characteristic three gives $(I+N{)}^3=I$. It is a nonabelian finite group of order $27$ whose nonidentity elements all have order three.

The group $PG{L}_2(F)=G{L}_2(F)/Z$ acts faithfully onby Möbius transformations. For $|F|=q$, this embeds $PG{L}_2(F)$ into the symmetric group $S_{q+1}$.

For any field $F$, a projective linear group element is uniquely determined by its images of three distinct points of the projective line, and any ordered distinct target triple is attainable. Given representatives $v_z,v_x$ forming a basis and $v_y=A{v}_z+B{v}_x$, the matrix with columns $A{v}_z,B{v}_x$ sends $(0,1,\infty )$ to $(x,y,z)$. Distinctness forces $A,B\ne 0$. The matrices realizing one projective map differ by a nonzero scalar; over ${\mathbb{F}}_p$ there are $p-1$ such matrices.

The projective line over a field $F$ is the set of one-dimensional subspaces of $F^2$ and may be identified with $F\cup \{\infty \}$.

The translations $x\mapsto x+b$ for $b\in {\mathbb{F}}_4$ form a Sylow $2$-subgroup of $PG{L}_2({\mathbb{F}}_4)$. Every nonidentity translation fixes infinity and exchanges the four finite points in two transpositions, so its action lies in $A_5$.

The determinant-one matrices form the normal subgroup $S{L}_n({\mathbb{F}}_q)=\ker (\det )$.

For a finite field of odd characteristic and nonzero $a,b$, let $U_a=\left(\begin{array}{cc}1& a\\ 0& 1\end{array}\right)$. An arbitrary determinant-one matrix $P=\left(\begin{array}{cc}r& s\\ t& u\end{array}\right)$ satisfies $P{U}_a=U_{b}P$ only if $t=0$ and $ar=bu$. The determinant condition $ru=1$ then gives $b/a=r^2$. Conversely $\mathrm{diag}(r,r^{-1})U_a\mathrm{diag}(r,r^{-1}{)}^{-1}=U_{a{r}^2}$. Thus nonzero upper unipotent matrices fall into two conjugacy classes under this square-class criterion. For example, $U_1$ and $U_3$ are conjugate over ${\mathbb{F}}_{11}$ because $5^2=3$, but not over ${\mathbb{F}}_5$. The criterion compares these particular upper unipotent representatives; it does not classify all determinant-one matrices.

A matrix $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ acts on the projective line by the Möbius transformation $x\mapsto (ax+b)/(cx+d)$, with the pole sent to infinity and infinity sent to $a/c$ when $c\ne 0$. The action is transitive because $\left(\begin{array}{cc}t& -1\\ 1& 0\end{array}\right)$ sends infinity to any finite $t$. The stabilizer subgroup of infinity consists of $\left(\begin{array}{cc}a& b\\ 0& a^{-1}\end{array}\right)$ with $a\ne 0$. The orbit-stabilizer theorem gives $|{\mathrm{SL}}_2({\mathbb{F}}_q)|=q(q^2-1)$. Its kernel is the scalar matrices of determinant one: a transformation fixing infinity has $c=0$, fixing zero then forces $b=0$, and fixing one gives $a=d$. Thus the kernel is $\{I,-I\}$, with these matrices coinciding in characteristic two.

The dihedral group $D_{2n}$ is generated by a rotation $r$ of order $n$ and a reflection $s$ with $srs=r^{-1}$; it has order $2n$.

For the dihedral group of a regular $2n$-gon, $n\ge 2$, use $r^{2n}=s^2=1$ and $srs=r^{-1}$. Its involutions are the central half-turn $r^n$ and all reflections $s{r}^j$. The half-turn has singleton conjugacy class and normal closure $\{1,r^n\}$. Conjugating a reflection changes its exponent by an even integer or negates it, so its conjugacy class is $\{s{r}^{j+2k}:0\le k<n\}$. Products of consecutive reflections generate $r^2$, giving normal closure $⟨r^2,s{r}^j⟩$, of order $2n$ and index two. The two reflection classes are distinguished by exponent parity.

If $|G|=p^{a}m$ with $p\nmid m$, then $G$ has subgroups of order $p^a$, every $p$-subgroup lies in one, and all such Sylow subgroups are conjugate. Their number satisfies $n_p\equiv 1(\mathrm{mod}p)$ and $n_p\mid m$.

If $H$ is a normal subgroup of a finite group $G$ and $P$ a Sylow subgroup of $H$, thenNormality puts every $g^{-1}Pg$ inside $H$, and Sylow conjugacy there gives $g^{-1}Pg=h^{-1}Ph$. Hence $g{h}^{-1}$ normalizes $P$, giving the factorization. This reduces questions about $G$ to a normaliser and its normal subgroup.

For distinct primes $p,q$, the Sylow theorems force a proper nontrivial normal subgroup in any group of order $p^{2}q$. If $p>q$, the Sylow $p$-subgroup is unique. If $p<q$ and both Sylow counts were nontrivial, the count congruences force $(p,q)=(2,3)$. In order twelve, four Sylow $3$-subgroups exhaust eight nonidentity elements; only three remain for a Sylow $2$-subgroup, making that subgroup unique. Thus the group is not a simple group.

A Sylow $p$-subgroup of a finite group has the largest power of $p$ dividing the group order.

Let $p>q$ be prime numbers. A nonabelian group of order $pq$ has a unique Sylow $p$-subgroup and exactly $p$ Sylow $q$-subgroups. Hence the Sylow count satisfies $p\equiv 1(\mathrm{mod}q)$, so $q\mid p-1$.

Up to isomorphism, the unique nonabelian group of order $21$ isIts nonidentity elements of $C_7$ form two conjugacy classes of size three, and the other two $C_7$ cosets form conjugacy classes of size seven.

A group, or one of its Sylow subgroups, acts on the set of Sylow subgroups by conjugation.

A finite nonabelian simple group embeds in $A_{n_p}$, where $n_p$ is its number of Sylow subgroups for a prime dividing its order. It is not a p-group by the nontrivial-centre property. Thus its Sylow subgroups are proper and cannot be normal, so $n_p>1$. The conjugation action is nontrivial, and its normal kernel of a group homomorphism must therefore be trivial. The sign of this faithful permutation action must be trivial, since a nontrivial sign homomorphism would inject the group into a group of order two. Lagrange's theorem consequently gives $|G|\mid n_p!/2$.

The Sylow $2$-subgroups of $S_3$ are its three transposition subgroups, while its unique Sylow $3$-subgroup is $A_3$. The three Sylow $2$-subgroups of $S_4$ are dihedral groups of order eight and are the normalizers of the three cyclic subgroups generated by inverse pairs of $4$-cycles. The five Sylow $2$-subgroups of $A_5$ are Klein four-groups, one fixing each letter.

If $P$ is a Sylow $p$-subgroup of $G$ and $Q$ is any $p$-subgroup, let $Q$ act on $G/P$. Since $p$ does not divide $|G/P|$, some coset $gP$ is fixed. Hence $g^{-1}Qg\le P$, or equivalently $Q\le gP{g}^{-1}$.

Suppose a group acts faithfully and transitively on seven points, every point stabilizer is $S_4$, and every two-point stabilizer is a Klein four-group. Then the group has order $168$ andCount pairs consisting of a point and a Sylow subgroup fixing it for $p=2,3$; the two-point stabilizer ensures uniqueness. For $p=7$, a normal Sylow subgroup would force the group into its order-$42$ normalizer in $S_7$.

Suppose a finite group $G$ has no index-two subgroup, $P$ is a Sylow $2$-subgroup, $H$ has index two in $P$, and $x\in G$ has order two. The sign of the action on $G/H$ defines a homomorphism $G\to \{\pm 1\}$ and is therefore trivial. Since $[G:H]\equiv 2(\mathrm{mod}4)$, a fixed-point-free involution would be an odd number of transpositions. Thus $x$ fixes a coset $gH$, equivalently $g^{-1}xg\in H$.

If a prime divides the order of a finite group, the group contains an element of that prime order.

One proof lets a cyclic group of order $p$ rotate the tuplesThe tuple set has size $|G{|}^{p-1}$, and its fixed tuples are precisely $(x,\dots ,x)$ with $x^p=1$. Counting nonfixed orbits modulo $p$ forces a nonidentity fixed tuple.

For a prime divisor $p$ of the order of a finite group, rotate the tuples $(g_1,\dots ,g_p)$ whose product is the identity. Rotation preserves that constraint even for a nonabelian group, by conjugating the product. Every nonfixed orbit has $p$ elements, while the fixed tuples correspond to $g^p=1$. There are $|G{|}^{p-1}$ tuples, so the number of fixed tuples is divisible by $p$. Since the identity is one, another solution exists and has order exactly $p$. This supplies an elementary proof of Cauchy theorem for groups.

For a finite group $G$, the map $x\mapsto x^m$ is bijective exactly when $m$ is coprime to $|G|$. If the two numbers are coprime, an inverse exponent modulo $|G|$ gives an inverse map. If a prime $p$ divides both, the Cauchy theorem for groups supplies a nonidentity element sent to the identity.

A composition series is a finite subnormal chain whose successive quotients are simple.

The composition length of a finite-length module or group is the number of simple factors in a composition series. The Jordan–Hölder theorem makes this independent of the chosen composition series.

A Jordan–Hölder factor is a simple quotient in a composition series. The Jordan–Hölder theorem says that the multiset of such factors is independent of the chosen composition series.

Any two composition series of a finite group or finite-length module have the same length and the same simple factors up to permutation and isomorphism.

The Klein four-group is the abelian group with three nonidentity elements, all of order two.