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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 $d$.

To adjoin an $n$th root of $g\in G$, let $m$ be the order of a group element $g$ and take a cyclic group $C=⟨a⟩$ of order $nm$ when $m$ is finite, or infinite order when $g$ has infinite order. In the free product $G\ast C$, $⟨a^n⟩$ and $⟨g⟩$ have the same order. The HNN extension with $t^{-1}{a}^{n}t=g$ therefore embeds its base by Britton's lemma, and $h=t^{-1}at$ satisfies $h^n=g$. If $g=1$, take $C$ of order $n$ and identify the two trivial subgroups. This is an existence construction, not an algorithm for finding an unknown element order.

Let $G$ have group order equal to a prime number $p$. Choose any $g\ne e$. By Lagrange's theorem, the order of a group element $g$ divides $p$, and it is greater than one. It is therefore $p$, so the cyclic subgroup generated by $g$ is all of $G$. This proves that every nonidentity element generates $G$, and $G$ is isomorphic to the additive group of integers modulo $p$.

For every positive integer $m$, $(g^{-1}{)}^m=(g^m{)}^{-1}$, hence $g^m=e$ if and only if $(g^{-1}{)}^m=e$. The sets of positive powers returning to the identity element are therefore equal. Taking the least positive exponent therefore gives the same finite order of a group element for $g$ and $g^{-1}$; if the sets are empty, both elements have infinite order. Equivalently, $g$ and $g^{-1}$ generate the same cyclic subgroup.

Two group elements $x$ and $y$ are conjugate when $y=gx{g}^{-1}$ for some group element $g$, equivalently when they belong to the same conjugacy class. Conjugate elements have the same order of a group element.

For every positive integer $n$, cancellation gives $(ha{h}^{-1}{)}^n=h{a}^n{h}^{-1}$. Thus the power on one side is the identity element exactly when $a^n=e$, proving preservation of the order of a group element, including infinite order. In particular, $ab$ and $ba$ are conjugate because $a^{-1}(ab)a=ba$, and hence have the same order.

In a finite group, a cyclic subgroup of order $m$ has exactly $\phi (m)$ generators, where $\phi$ is the Euler totient function. Each element generates exactly one cyclic subgroup, so dividing the number of elements of order $m$ by $\phi (m)$ counts these subgroups. Include the identity subgroup using $m=1$ and $\phi (1)=1$. In a symmetric group, element counts come from cycle type and order of a group element is the least common multiple of cycle lengths.

The Coxeter number is the order of a Coxeter element in a finite irreducible Coxeter group.

The finite cyclic group $C_4=\mathbb{Z}/4\mathbb{Z}$ and the Klein four-group $V_4=C_2\times C_2$ are both abelian groups of order four. In $C_4$, composition is addition modulo four. In $V_4$, label the elements $00,01,10,11$ and add each coordinate modulo two. For example, $01+10=11$ and $11+11=00$.

The order of a group element is preserved by a group isomorphism: an equation $g^m=e$ holds exactly when its image satisfies the same equation. In $C_4$, both $1$ and $3$ have order four. In $V_4$, each nonidentity element has order two. Hence these groups are not isomorphic, despite having the same order of a finite group. No relabelling can repair this discrepancy.

For a geometric model of $V_4$, take the symmetries of a non-square rectangle: the identity element, reflection in its horizontal axis, reflection in its vertical axis, and rotation through $180^{\circ }$. The two reflections commute, and their composite is the half-turn. Every nonidentity motion is its own inverse element. This makes the two binary switches concrete: choosing whether to apply each reflection specifies all four motions. The case of abelian groups of prime-square order generalizes this distinction from $p=2$ to any prime $p$.

For positive integers $m,n$, two finite cyclic groups combine into one cyclic group precisely when their group orders are coprime:The reason is that both cyclic counters must return to their identity elements simultaneously. If $a,b$ are generators of a group for the two factors, the order of a direct-product element gives $\mathrm{ord}(a,b)=\mathrm{lcm}(m,n)$. The pair generates all $mn$ elements exactly when this least common multiple is $mn$.

For $C_2\times C_3$, repeated addition of $(1,1)$ visitsAll six pairs appear before the return, so $(1,1)$ is a generator. In contrast, in $C_2\times C_2$ the same step gives only $(0,0),(1,1),(0,0)$. Every nonidentity element has order of a group element two, so none can generate all four pairs. This is the Klein four-group; there is no group isomorphism to $C_4$ even though both have four elements.

For a brief proof of the converse, each component order of a group element divides the corresponding factor group order. Every pair's order therefore divides $\mathrm{lcm}(m,n)$. If $\gcd (m,n)>1$, this is smaller than $mn$, so no pair can generate the whole product. The Chinese remainder theorem gives the explicit isomorphism in the coprime case: $t\mathrm{mod}mn\mapsto (t\mathrm{mod}m,t\mathrm{mod}n)$.

The cyclic subgroup generated by an element $g$ is $⟨g⟩=\{g^n:n\in \mathbb{Z}\}$. Its size is the order of $g$.

Let $G$ be a finite group and $\varphi :G\to H$ a surjective group homomorphism. Choose a preimage $g$ of $h$. Since $\varphi (g{)}^{o(g)}=e$, the order of a group element $m=o(h)$ divides $o(g)$; write $o(g)=mr$. Then $g^r$ has order $m$, because $(g^r{)}^s=e$ holds exactly when $mr\mid rs$, or equivalently $m\mid s$. This proves existence of an element of order $m$ in $G$. It need not be a preimage of $h$: for the quotient map ${\mathbb{Z}}_4\to {\mathbb{Z}}_2$, both preimages of the nonidentity target element have order $4$, while the order-$2$ element of the source maps to zero. Thus the assertion concerns element-order existence, not splitting of the quotient map.

The exponent of a finite group is the least common multiple of the orders of all its elements.

Every finite group is a torsion group; its exponent is a common multiple of the order of a group element for every element.

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.

A group element has finite order if some positive integer power equals the identity element. Its order of a group element is the least such positive integer.

A group automorphism is a bijective group homomorphism $\alpha :G\to G$. It preserves the group operation, the identity element, and inverse elements; its inverse is another group homomorphism. Thus it is a group isomorphism with identical source and target. These maps form the automorphism group $\mathrm{Aut}(G)$ under function composition.^[ref]

For the additive finite cyclic group $C_6$, take $f_5(k)=5k(\mathrm{mod}6)$. Its output list on $0,1,2,3,4,5$ is $0,5,4,3,2,1$, a permutation of all six elements. It preserves addition because $5(a+b)=5a+5b(\mathrm{mod}6)$, and applying it twice gives $25k=k(\mathrm{mod}6)$. Hence it is a group automorphism and its own inverse function. Compare the collapsing $f_2$ in the figure under endomorphism. The general criterion under endomorphism of a finite cyclic group says exactly when such a map is reversible: its multiplier must be coprime to the size of the finite cyclic group.

A permutation need not be a group automorphism, even if it fixes the identity element. Let $\pi$ exchange $1$ and $2$ in $C_6$ and fix $0,3,4,5$. It is a bijection, and $\pi (0)=0$, butThus it fails to preserve addition and is not even a group homomorphism. Reversibility alone does not preserve the group operation.

A group automorphism carries each cyclic subgroup onto the cyclic subgroup generated by the image of its defining element, preserving the order of a group element. In particular, it carries a generator of a group to another generator of a group. For $C_6$, the existing calculation of generators of a finite cyclic group leaves only the images $1$ and $5$ for $1$, so its automorphism group consists exactly of the identity function and $f_5$. This turns the problem of compatible relabeling into a much smaller choice than enumerating arbitrary permutations.

A group isomorphism is a bijective group homomorphism $\phi :G\to H$. It matches the entire group operation: $\phi (ab)=\phi (a)\phi (b)$, with each product taken in its own group. Its inverse is automatically a group homomorphism. A group automorphism is the special case $H=G$.^[ref]

The target is part of the claim. The doubling example under isomorphism is a group isomorphism $d:\mathbb{Z}\to 2\mathbb{Z}$, between the additive integers and the additive even integers. Exactly the same rule $d(k)=2k$, declared as $d:\mathbb{Z}\to \mathbb{Z}$, is an injective endomorphism that misses $1$ and every other odd integer, so is not a group automorphism. Changing the declared target changes surjectivity, even though it does not change the formula or its compatibility with addition.

An isomorphism lets a calculation in one model solve the same problem in another. The example under group identifies addition modulo four with function composition of rotations of a square via $k\mapsto r^k$. Through this group isomorphism, adding exponents corresponds exactly to composing motions. More generally, a group isomorphism preserves order of a group element, carries subgroups to subgroups, and carries a generating set of a group to a generating set of a group. Having equally many elements is insufficient: the permutation in group automorphism already fails compatibility on two inputs.

Why does the inverse preserve the group operation? Write $u=\phi (a)$ and $v=\phi (b)$, possible because of surjectivity. ThenThis proves the inverse condition directly. Analogous calculations work for each operation of a ring and for addition and scalar multiplication in a vector space. The argument relies on preserving those operations; it supplies no general continuity claim for the inverse of a continuous bijection between topological spaces.

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.

For any group $G$, evaluation at a chosen generator of a group $x$ of $C_n$ gives a bijection between group homomorphisms $C_n\to G$ and elements satisfying $g^n=1$. A group homomorphism must send $x^j$ to $g^j$. Conversely, this formula is well-defined because exponents differing by a multiple of $n$ give equal powers, and it respects multiplication. The displayed correspondence is a bijection of sets; it is not claimed to be a group isomorphism for a nonabelian target. In a symmetric group, the order of a group element is the least common multiple of its disjoint cycle lengths, making this correspondence easy to enumerate.

The ring of integers is $\mathbb{Z}[\sqrt{35}]$ and the field discriminant is $140$. The Minkowski bound for ideal classes is $\sqrt{35}<6$. At three the defining polynomial is irreducible, so the only prime ideals needed to generate the ideal class group are $P_2=(2,\sqrt{35}-1)$ and $P_5=(5,\sqrt{35})$, of ideal norms two and five. They satisfy $P_2^2=(2)$, $P_5^2=(5)$ and $P_2{P}_5=(5+\sqrt{35})$, so one class of order of a group element at most two generates. The field norm equation $a^2-35b^2=\pm 2$ is impossible modulo five, proving $P_2$ is not a principal ideal. Hence the ideal class group has exactly two elements.