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Every group of order six is isomorphic to the cyclic group $C_6$ or the symmetric group on three points $S_3$. Lagrange's theorem restricts element orders to $1,2,3,6$. If there is an element of order six, it generates the group. If every nonidentity element had order two, the group of exponent two is abelian result would produce a subgroup $\{e,a,b,ab\}$ of order four, contradicting Lagrange's theorem. Thus the noncyclic case has an element $x$ of order three.

There cannot be two distinct subgroups of order three: their intersection would be trivial and their nine pairwise products would be distinct elements of a six-element group. Hence exactly two elements have order three, and the other three nonidentity elements have order two. The subgroup $⟨x⟩$ has index two and is normal. Choose an involution $h$ outside it. Conjugation sends $x$ to $x$ or $x^{-1}$. The first case would make $xh$ have order six, so $hx{h}^{-1}=x^{-1}$. The generators therefore obey $x^3=h^2=e$, $hx{h}^{-1}=x^{-1}$, the relations realized by $(123)$ and $(12)$ in $S_3$.

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 group of exponent two is abelian. Give such a finite group the additive notation of a vector space over ${\mathbb{F}}_2$, with zero acting as zero and one as the identity scalar. The group axioms and $g+g=0$ verify the vector-space laws. A basis with $d$ vectors then gives exactly $2^d$ elements. Thus a finite group whose elements all have order at most two must have power-of-two order.

Let a finite group $G$ act on a set $X$. For $x\in X$, define its stabilizer subgroup $G_x=\{g:g\cdot x=x\}$ and its group orbit $Gx=\{g\cdot x:g\in G\}$. The mapis well-defined, because multiplying $g$ on the right by an element fixing $x$ does not change its image. It is onto by the definition of the orbit. If $g\cdot x=h\cdot x$, then $h^{-1}g\in G_x$, so $g{G}_x=h{G}_x$; hence it is one-to-one. Every coset has $|G_x|$ elements and the cosets partition $G$. This proves the orbit-stabilizer theorem:For a group element $x$, its cyclic subgroup $H=⟨x⟩$ has $\mathrm{ord}(x)$ elements. Apply the theorem to the action of $G$ on its left cosets $G/H$, whose stabilizer at $H$ is $H$. It follows thatThis is the needed case of Lagrange's theorem, obtained directly from the action.

To prove Cauchy's theorem for finite groups, let $p$ be a prime divisor of $|G|$ and considerThe first $p-1$ entries are arbitrary and determine the last, so $|\mathcal{X}|=|G{|}^{p-1}$ is divisible by $p$. Cyclically rotate the entries. This preserves the product constraint even if $G$ is nonabelian, sinceThus a cyclic group of order $p$ acts on $\mathcal{X}$. By the orbit-stabilizer theorem, its orbits have size one or $p$. Its fixed tuples are exactly $(g,\dots ,g)$ with $g^p=e$. If $m$ is their number, counting the remaining size-$p$ orbits gives $m\equiv |\mathcal{X}|\equiv 0(\mathrm{mod}p)$. The identity provides one fixed tuple, so $m\ge p$ and a nonidentity $g$ with $g^p=e$ exists. The order of this $g$ divides $p$: division of $p$ by its order and minimality of that order prove this directly. Since $p$ is prime and $g\ne e$, the order is exactly $p$. This is the cyclic-tuple proof of Cauchy theorem.

Now suppose every nonidentity element has order two. Then every element equals its inverse, and for any $a,b\in G$,Thus group of exponent two is abelian applies. Starting from the identity subgroup, choose an element outside the subgroup already generated. Because the group is abelian and that element squares to the identity, adjoining it doubles the subgroup: it is the disjoint union of the old subgroup and its translate by the new element. Repeating until the finite group is exhausted shows $|G|=2^r$. Conversely the direct product of groups $C_2^r$ has order $2^r$ and every nonidentity element has order two. ThereforeThe case $r=0$ is the trivial group and satisfies the condition vacuously. These finite examples are elementary abelian groups of exponent two.

An infinite example is the group of all binary sequences, with coordinatewise addition modulo two:Every sequence added to itself is zero, so every nonzero sequence has order exactly two. The sequences having a single $1$ in coordinate $j$ are all distinct, proving that the group is infinite.

For a finite group $G$ and a subgroup $H$, Lagrange's theorem statesIn particular $|H|$ divides $|G|$: the left cosets partition $G$, and multiplication by a representative is a bijection from $H$ to each coset. Apply this to the cyclic subgroup $⟨g⟩$. Its size is the order of a group element $g$, so every element order divides $|G|$.

Under the square condition, every group element is its own inverse element. Consequentlyfor all $g,h$. This proves that a group of exponent two is abelian, including the trivial group.

For the fourth-power condition a counterexample to commutativity is the quaternion groupHere $i^2=j^2=k^2=-1$ and $ij=k=-ji$. Thus its six elements outside $\{\pm 1\}$ have order of a group element four, while $1$ and $-1$ have orders one and two. All fourth powers are the identity, yet $ij\ne ji$.

No such nonabelian group exists. Every element, including the identity, satisfies $g^2=1$, so $g^{-1}=g$. For any $a,b$,Therefore the group is an abelian group. This proves that a group of exponent two is abelian without any finiteness assumption.

The hypothesis implies $x^{-1}=x$ for every $x\in G$, including $e$. For $a,b\in G$,Thus $G$ is an abelian group. This is the group of exponent two is abelian argument.

The converse is false: the cyclic group $C_3$ is an abelian group, but its two nonidentity elements have order of a group element three.

A left coset is $gH=\{gh:h\in H\}$. The cosets partition a finite group into equally sized sets of $|H|$ elements, so Lagrange's theorem gives $|G|=[G:H]|H|$; in particular subgroup order divides group order.

Here $A=-I$, $C=-B$ and $B^2=I$. The Cayley table, with row element multiplied on the left of the column element, is

The table proves closure, $I$ is the identity and every element is its own inverse. Associativity is inherited from matrix multiplication, so this is the Klein four-group.

Finally, if every nonidentity element has order two, every element satisfies $h^{-1}=h$, including the identity. Then $ab=(ab{)}^{-1}=b^{-1}{a}^{-1}=ba$. Thus a group of exponent two is abelian.

A group has a closed binary operation, associativity, an identity, and an inverse for each element. For a finite group, Lagrange's theorem states $|G|=[G:H]|H|$ for each subgroup $H$, so $|H|$ divides $|G|$.

The order of a group element $g$ is the least positive $n$ with $g^n=e$, or infinity if none exists. In a finite group the powers form the cyclic subgroup $⟨g⟩$ of that order; Lagrange's theorem proves $\mathrm{ord}(g)\mid |G|$. If $|G|=p$ is prime and $g\ne e$, its order must be $p$. Thus $g$ generates the whole group, proving $G\cong C_p$.

If every nonidentity element has order two, every element is its own inverse. Then $ab=(ab{)}^{-1}=b^{-1}{a}^{-1}=ba$. This proves that a group of exponent two is abelian.

Lagrange's theorem states that the group order of a finite group divides the group order; more precisely $|G|=[G:H]|H|$.

If every element squares to the identity element, every element is its own inverse element, and $ab=(ab{)}^{-1}=b^{-1}{a}^{-1}=ba$. Thus a group of exponent two is abelian. For distinct nonidentity $a,b$, the four elements $1,a,b,ab$ are distinct: for example $ab=1$ would imply $a=b$, and $ab=a$ would imply $b=1$. Products remain in this set because $a,b$ commute and square to one. Thus they form a Klein four-group subgroup of order four.

Now let $|G|=2p$ with odd prime number $p$. By Lagrange's theorem, every element order is one of $1,2,p,2p$. If an element has order $2p$, its square has order $p$. If no element had order $p$, all nonidentity elements would therefore have order two. There are at least two distinct such elements, producing a subgroup of order four. This contradicts $4\nmid 2p$. Hence there is an element of order $p$.