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An abelian group presentation specifies generators of a group and additive relations. For generators $e_1,\dots ,e_n$, let $L$ be the subgroup of ${\mathbb{Z}}^n$ generated by the relation vectors; the presented abelian group is ${\mathbb{Z}}^n/L$. The orders of the images of the generators can be smaller than the orders imposed by individual relations. A relation $d{e}_i=0$ only requires that the order of a group element divide $d$.

For odd $n\ge 3$, the dihedral group $D_{2n}$ of order $2n$ has rotation generator $r$ and reflection generator $s$, with $srs=r^{-1}$. The group commutator $[r,s]=r^{-2}$ generates the rotation subgroup, while the quotient by that subgroup is $C_2$. Hence the commutator subgroup is exactly $⟨r⟩$ and the abelianization is $C_2$. The only abelian quotients are $C_2$ and the trivial group.

In the Heisenberg group over a commutative ring with $R=\mathbb{Z}/4\mathbb{Z}$, the subset $H=\{A(t,y,t):t,y\in R\}$ is an Abelian subgroup of group order 16. Its product has central coordinate $y+y^{\prime }+t{t}^{\prime }$, symmetric in the two factors. Its inverse element is $A(-t,t^2-y,-t)$. The map $A(x,y,z)\mapsto x-z$ is a surjective group homomorphism to the additive group of $R$, with kernel of a group homomorphism $H$. Thus $H$ is a normal subgroup and has four cosets in the 64-element group.

Unlike the odd-prime case, division by two is unavailable. In fact take $u=A(1,0,1)$ and $v=A(2,0,2)$. Then $u^n=A(n,(\genfrac{}{}{0ex}{}{n}{2}),n)$, where the binomial coefficient is computed as an integer before reduction modulo four. Hence $u^4=A(0,2,0)$ and $u$ has element order eight. The element $v$ has element order two and is not a power of $u$: the powers with outer coordinates two have central coordinates one or three. Their generated subgroups intersect trivially, so the 16-element Abelian group is the direct product of groups $C_8\times C_2$.

An abelian subgroup is a subgroup that is an abelian group under the restricted operation.

If an abelian subgroup $A$ has index $p^a$, the centralizer of any $1\ne g\in A$ contains $A$, so the class size of $g$ divides $p^a$. The prime-power conjugacy-class obstruction to simplicity rules this out. If $A=1$, the group is a p-group and has nontrivial center of a group, also ruling out nonabelian simplicity.

Fix a prime number $p$. Take the full subcategory of finitely generated abelian groups with no element of order $p^2$. It is closed under finite direct sums and subgroups. If $t_p(G)$ is the $p$-primary torsion subgroup, the map $G\to R(G)=G/(p{t}_p(G))$ is a reflection into this subcategory: every map to an allowed group kills $p{t}_p(G)$. Therefore ordinary kernels in a category and reflected cokernels in a category supply all finite categorical limits and colimits. No nonzero finitely generated group has zero reflection, so the epimorphisms are precisely surjections and are all conormal epimorphisms. Multiplication by $p^2$ on $\mathbb{Z}$ is a monomorphism but not a normal monomorphism: its cokernel here is $\mathbb{Z}/p\mathbb{Z}$, whose kernel is $p\mathbb{Z}$, not $p^2\mathbb{Z}$. Multiplication by $p$ is normal; composing it with itself gives this nonnormal monomorphism.

The additive energy of a finite set $S\subseteq G$ in an abelian group $G$ counts additive quadruples, with order and repeated members allowed:At a fixed sum $s$, independently choosing two of its $r_{S+S}(s)$ ordered pairs gives $r_{S+S}(s{)}^2$ additive quadruples. Thus additive energy weights heavily represented sums more strongly than the cardinality of the sumset does.

For $A=\{0,1,2,3\}$ and $B=\{0,1,4,10\}$, the sumset representation functions giveThe arithmetic progression has additional coincidences such as $0+3=1+2$. The set $B$ still has coincidences forced by commutativity, including $0+1=1+0$, so its additive energy is not merely the sixteen identical-pair comparisons.

The same additive energy can be counted using the difference set. Define $r_{S-S}(t)=|\{(a,b)\in S\times S:a-b=t\}|$. The identity $a+b=c+d$ is equivalent to $a-c=d-b$, and the coordinate rearrangement $(a,b,c,d)\mapsto (a,c,d,b)$ is a bijection. ConsequentlyFor $A$, the difference counts from $-3$ through $3$ are again $1,2,3,4,3,2,1$. For $B$, the zero difference has four representations and each of the twelve nonzero differences has one, giving $4^2+12\cdot 1^2=28$. The sumset representation functions for sums and differences can look different while their squared totals agree.

For every nonempty finite set $S\subseteq G$ in an abelian group $G$,The lower bound follows from the Cauchy-Schwarz inequality and the total sumset representation function count:For the upper bound, any three members $a,b,c$ determine at most one fourth member $d=a+b-c$ of an additive quadruple. See Yufei Zhao's MIT lecture 25, pages 1-3, for representation counts and the energy bounds.

In the two integer examples, the lower bounds are $256/7$ and $256/10$, and the upper bound is $64$. For the subgroup $H=\{0,2,4\}$ of $\mathbb{Z}/6\mathbb{Z}$, each element of $H$ has three sum representations, so $E(H)=3\cdot 3^2=27$ and both bounds are equalities.

For nonempty $S$, dividing additive energy by $|S{|}^3$ gives a relative measure between $1/K(S)$ and one. Small doubling constant therefore forces relatively large additive energy. A converse requires care: the Balog-Szemerédi-Gowers theorem extracts a large subset with small doubling constant, rather than asserting that all of $S$ has small doubling constant.

The integers form an infinite cyclic group under addition, with identity element zero and inverse element $-n$ for $n$. Its two generators are $1$ and $-1$. Every subgroup is $m\mathbb{Z}$ for a unique nonnegative integer $m$: if it is nonzero, take its least positive element and use division with remainder to show that every element is a multiple of it. The zero subgroup corresponds to $m=0$.

The Adjoint representation of G2 has one weight at each of its twelve roots and two at zero. Its six long roots form the A2 root system; together with the two zero weights they give the Adjoint representation of SU(3). The remaining six short roots form the weight systems of its defining representation and dual representation. Hence restriction to the long-root SU(3) gives $14=8\oplus 3\oplus \overline{3}$. This is a branching of a representation; the complement is invariant under the subgroup, not under the full G2 action.

The Sylow theorems give either one or eight Sylow subgroups of order seven. In the first case that subgroup is normal. In the second, their six nonidentity elements are pairwise disjoint, accounting for 48 elements. Only eight elements remain, including the identity, and every Sylow subgroup of order eight lies entirely among them. Thus there is exactly one, and it is a normal subgroup.

Cayley theorem embeds a finite group of order $n$ into $S_n$. Extend $\sigma \in S_n$ to $n+2$ letters and multiply it by the disjoint transposition $(n+1n+2)$ exactly when $\sigma$ is odd. Additivity of parity makes this an injective group homomorphism into the alternating group $A_{n+2}$. Thus every finite group is isomorphic to a subgroup of some alternating group.

For a subgroup $K$ of a finite abelian group $G$, its annihilator isThese are exactly the linear characters that descend to the quotient group $G/K$. The character group of a finite abelian group has the same size as the group, so $|K^{\perp }|=|G/K|=|G|/|K|$. The annihilator is what abelian hidden-subgroup Fourier sampling measures.

For a finite simple group with a subgroup of index $n\ge 3$, the coset action is nontrivial and its kernel is a normal subgroup, hence trivial. Its sign homomorphism must also be trivial: otherwise simplicity would give an isomorphism to $C_2$, which cannot act transitively on at least three points. Thus the image lies in the alternating group. In particular Lagrange's theorem gives $|G|\mid n!/2$. This can exclude simplicity with very little counting: a group of order $112$ has a subgroup of index seven by Sylow theorems, but $112$ does not divide $7!/2$.

Every subgroup of an amenable group is amenable. One proof restricts an invariant mean after extending bounded functions from the subgroup to the whole group along a set of coset representatives. Consequently, a group containing a nonamenable subgroup is nonamenable.

Choose distinct odd primes and take products of the nonisomorphic Gassmann equivalent regular subgroups, selecting one of the two groups at each prime. Product conjugacy-class counts give $2^n$ Gassmann equivalent subgroups of the same finite product of symmetric groups. Their fundamental groups are distinguished by whether the unique Sylow subgroup at each chosen prime is abelian. The closed-manifold realization of a finitely presented fundamental group and Sunada theorem therefore produce $2^n$ pairwise nonhomeomorphic, and hence nonisometric, isospectral closed four-manifolds. Gassmann equivalence alone, without a topology distinction, does not imply nonhomeomorphism.

The branching of a group representation $V$ to a subgroup $H$ is its decomposition into irreducible representations of $H$, when that restriction of a representation is a semisimple representation: $V{|}_H={⨁}_{\lambda }{m}_{\lambda }{W}_{\lambda }$. The nonnegative integers $m_{\lambda }$ are the branching multiplicities. For finite groups over the complex numbers, they are the character inner products of the restricted character with the irreducible characters of $H$. Restriction to an arbitrary subgroup need not be completely reducible.

A branching rule decomposes a representation into irreducible representations after restriction from a Lie group or Lie algebra to a subgroup or subalgebra.

A branching rule gives the irreducible representation decomposition after restricting a group representation to a subgroup, or a Lie algebra representation to a Lie subalgebra. For the standard block embedding $SU(5)\subset SU(6)$, $6\downarrow =5\oplus 1$ and ${\Lambda }^{2}6\downarrow =10\oplus 5$.

Let $W$ be the subgroup fixed by the Coleman norm operator. The Coleman norm-fixed lift gives the unique factorization $f=w(\overline{f})v$, with $v\in 1+pR$. Multiplication consequently gives the displayed direct product of groups, while reduction identifies $W$ with ${\mathbb{F}}_p[[T]{]}^{\times }$. This is a decomposition of unit groups, not an identification of a power-series ring with a product of groups.

For the additive infinite cyclic group $\mathbb{Z}$, the generating set of a group $S=\{-1,1\}$ produces the infinite line: each integer $n$ is adjacent to $n-1$ and $n+1$. Adding the moves $\pm 2$ gives $T=\{-2,-1,1,2\}$ and additional edges skipping one vertex. Each skip still costs one step in the word metric.

For integers $a,b$, these word metrics areThe absolute value counts unit moves, and the ceiling function counts pairs plus a possible leftover unit. Every $T$-move changes the integer by at most two, while moves of size two followed by at most one move of size one attain this bound. For example, the graph path $0,1,2,3,4,5$ has five steps for $S$, while $0,2,4,5$ has three for $T$. This calculation gives the concrete comparison

The picture shows only a finite window of each infinite Cayley graph. By contrast, $U=\{-2,2\}$ alone fails to be a generating set of a group for $\mathbb{Z}$: it generates the subgroup $2\mathbb{Z}$. Its move graph has two connected components of a graph, the even integers and the odd integers. Within either component the graph distance is $|a-b|/2$; between different components no finite graph path exists. Adding a shortcut is different from deleting the moves needed for connectivity.

Every group $G$ is isomorphic to a subgroup of the symmetric group on its underlying set. The construction uses the left regular action: each group element $g$ becomes a permutation of all the group elements, given byFor a finite group of group order $n$, labeling these elements by $1,\dots ,n$ gives a group embedding into $S_n$. The same construction works for an infinite group, with the symmetric group of all bijections of its underlying set.^[ref]

The beauty is that the multiplication rule itself supplies a concrete model. Whatever the group elements originally represent, each one rearranges the entire group by multiplication. A product becomes function composition, and no distinctions between group elements are lost. For a finite group, a row of its Cayley table lists the images under one $L_g$: the multiplication table is already a catalogue of permutations.

Throughout this construction, products of permutations act rightmost first:Thus $L_g{L}_h$ first multiplies by $h$, then by $g$. This is a left group action, not a license to reverse the multiplication order in a nonabelian group. Ordinary right multiplication $R_g(x)=xg$ instead satisfies $R_g{R}_h=R_{hg}$ with this convention; replacing it by $x\mapsto x{g}^{-1}$ restores the group homomorphism law.^[ref]

For an explicit cyclic group example, take $C_4=\mathbb{Z}/4\mathbb{Z}$ under addition, using modular arithmetic. Its left regular action consists of $L_k(x)=k+x(\mathrm{mod}4)$. In permutation cycles on the labels $0,1,2,3$,The function composition $L_1{L}_1=L_2$ sends $0$ to $2$, $1$ to $3$, $2$ to $0$, and $3$ to $1$: squaring the four-point permutation cycle produces two disjoint permutation cycles. Also $L_1{L}_3=L_0$. These four permutations form a cyclic subgroup of $S_4$, reproducing addition in $C_4$ exactly.

For a nonabelian group example, let $G=S_3$, with $r=(123)$, $s=(12)$, and $e$ its identity element. The six group elements are $e,r,r^2,s,rs,r^{2}s$, where $r^3=s^2=e$ and $sr=r^{2}s$. Now the labels being permuted are themselves permutations of three points. Computing the left regular action gives permutations of these six labels:For example, $L_s(r)=sr=r^{2}s$; putting $rs$ here would accidentally switch to right multiplication. Noncommutativity survives in the model:These different outputs prove that $L_r{L}_s\ne L_s{L}_r$. The group generators $r,s$ become group generators $L_r,L_s$ of a six-element subgroup of $S_6$.

Why does a group action here produce a group embedding? Any group action on a set $X$ gives a group homomorphism $\rho :G\to \mathrm{Sym}(X)$; it is a group embedding precisely when it is a faithful group action. The group kernel consists of the group elements fixing every point of $X$. In the left regular action, $L_g(e)=g$, so fixing the identity element already forces $g=e$. In fact every stabilizer subgroup is trivial, since $gx=x$ forces $g=e$; this stronger property makes the left regular action a regular group action.^[ref]

Compare two group actions of the same cyclic group $C_4$. Its left regular action on four labels distinguishes all four group elements. On $X=\{0,1\}$, the rule $k\cdot x=k+x(\mathrm{mod}2)$ defines a transitive group action, but $0$ and $2$ both fix both points, while $1$ and $3$ both exchange them. Its group kernel is $\{0,2\}$, so it is not a faithful group action. Its image is a two-element subgroup of $S_2$, isomorphic to the quotient group $C_4/\{0,2\}$ rather than to $C_4$.

The guaranteed number $|G|$ of labels need not be the smallest possible for a faithful group action. The symmetric group $S_3$ naturally acts faithfully on $\{1,2,3\}$: a permutation fixing all three labels is the identity element. Its left regular action uses six labels instead. Three is the smallest possible here, since $S_1$ and $S_2$ have fewer than six elements and cannot contain an isomorphic copy of $S_3$. Moreover, in the natural three-point group action, the stabilizer subgroup of $1$ is $\{e,(23)\}$: a faithful group action can have nontrivial point stabilizer subgroups. Cayley's construction supplies a universal model, while a smaller model can reveal particular symmetries.

The proof is short. Each $L_g$ is a bijection with inverse $L_{g^{-1}}$. Associativity givesso $\lambda$ is a group homomorphism. If $L_g=L_h$, evaluating both at the identity element gives $g=L_g(e)=L_h(e)=h$. Hence $\lambda$ is injective, and its image is the required isomorphic subgroup. No finiteness assumption entered the argument.

The historical connection is unusually close to the notation. The opening paragraph of Arthur Cayley's 1854 paper already treats permutations as one example of operations on a collection of quantities and explicitly performs the rightmost operation first in a product. It also allows noncommutativity while insisting on associativity. The modern left regular action shows how that viewpoint reaches every abstract group, by choosing the group itself as the set to be acted on.^[ref]