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In characteristic $2$, a Sylow subgroup $P$ of $A_5$ is a Klein four-group, and its normalizer is $A_4$. Since $k{A}_4$ has one block, Brauer first main theorem gives a unique block of $k{A}_5$ with defect $P$. No order-two subgroup $D$ can be a defect group of a block: both $C_{A_5}(D)$ and $C_{A_5}(P)$ equal $P$, so the Brauer morphisms of a central block idempotent at $D$ and at $P$ are equal. A nonzero image is $1$, since a group algebra of a p-group in characteristic p is local, and consequently $D$ is not maximal among subgroups with nonzero Brauer image.

An abelian group is a group whose group operation is commutative: $gh=hg$ for every $g,h\in G$. Thus rearranging a finite sequence of operations preserves the result. Associativity already permits changing parentheses; commutativity additionally permits exchanging the elements.

The additive integers, the additive real numbers, and every cyclic group are abelian groups. Additive notation is often convenient: the identity element is written $0$, the inverse element of $a$ is $-a$, and $a+b=b+a$. It does not force an element to be an ordinary number: in $C_2\times C_2$, an element is a pair of binary coordinates, added componentwise modulo two.

Commutativity is useful because an integer linear combination of chosen group elements depends only on their coefficients, rather than on the order in which the operations were performed. The Fundamental theorem of finitely generated abelian groups turns this simplicity into a classification by cyclic groups. However, commutativity does not imply that a single generator of a group suffices: the Klein four-group is the smallest counterexample, as the cyclic group of order four and Klein four-group demonstrate.

Let $D_8$ denote the dihedral group of group order eight acting on a square. Here the subscript specifies group order; the same group is called $D_4$ in the convention that counts vertices of a polygon, as in the group action example. Let $r$ be a counterclockwise quarter-turn rotation and $s$ a reflection in a diagonal. ThenProducts act rightmost first. The group commutator of the generators of a group isIt is a half-turn rotation, so forcing $r$ and $s$ to commute must identify a half-turn with $e$.

More concretely, write each element uniquely as $r^i{s}^j$, with $0\le i<4$ and $j\in \{0,1\}$. The functionis a surjective group homomorphism. Indeed, $srs=r^{-1}$ giveswhere the exponents are reduced modulo four and two. Modulo two, $-k$ and $k$ agree, so $\varphi$ adds the two coordinates separately. Its group kernel is $N=\{e,r^2\}$. Since the target is an abelian group, $D_8^{\prime }\subseteq N$; since $[r,s]=r^2$, also $N\subseteq D_8^{\prime }$. ThusThe four cosets are $\{e,r^2\}$, $\{r,r^3\}$, $\{s,r^{2}s\}$, and $\{rs,r^{3}s\}$. Each coset forgets the difference of a half-turn. The quotient group still distinguishes rotations from reflections, and even from odd numbers of quarter-turns; the two surviving choices are independent. This is the Klein four-group, rather than the cyclic group $C_4$, because each nonidentity element squares to the identity element. For the general dihedral group calculation, see Conrad, Dihedral groups I, theorem A.1; that source uses the vertex-count convention and $xy{x}^{-1}{y}^{-1}$ for individual group commutators.

Write the Klein four-group as $\{1,a,b,ab\}$. Its four one-dimensional irreducible representations independently assign signs to $a,b$, so the displayed characters have $p,q\in \{0,1\}$. The character table isEvery conjugacy class is a singleton because the group is abelian. The four rows exhaust the complex irreducibles by the equality between their number and the number of classes. If $H$ is this table, $H{H}^T=4I$ proves character orthogonality. The diagonal three-dimensional representation by half-turns about the coordinate axes is the direct sum of the three nontrivial characters, with total character $(3,-1,-1,-1)$.

Let $B(x)=\left(\begin{array}{cc}\cosh x& -\sinh x\\ -\sinh x& \cosh x\end{array}\right)$, $T=\mathrm{diag}(1,-1)$ and $P=\mathrm{diag}(-1,1)$. Every real matrix preserving $\mathrm{diag}(1,-1)$ has a unique form $B(x)S$ with $S\in \{I,T,P,TP\}$. The boost subgroup is normal because $TB(x)T=PB(x)P=B(-x)$. Its four cosets form the Klein four-group, giving the displayed quotient group. In contrast, the four-element subgroup generated by $T,P$ is not normal: $B(x)TB(-x)=B(2x)T$ is not diagonal for $x\ne 0$. This distinguishes the normal identity component from a chosen set of component representatives.

For the convention in cross-ratio normalization of four projective points, permuting four distinct projective points produces the six displayed rational expressions. Interchanging the first pair sends $\lambda$ to $1/\lambda$; interchanging the middle pair sends it to $1-\lambda$. The four permutations consisting of the identity and the three double transpositions form a Klein four-group preserving the cross-ratio. The resulting six-element action gives every displayed expression.

For real $\lambda \notin \{0,1\}$, repeated values occur exactly for $\lambda \in \{-1,1/2,2\}$, and then the orbit has size three. To prove this, a repetition gives a nonidentity stabilizer after composing one operation with the other's inverse. Its fixed-point equations give $-1$, $1/2$, $2$, or ${\lambda }^2-\lambda +1=0$, with the last case nonreal. With one point at infinity, $\lambda =(x_1-x_3)/(x_1-x_2)$; the three real exceptional values mean that one finite point is the midpoint of the other two. Thus these harmonic conjugate points configurations have three collinear equally spaced finite points in that affine coordinate.

The full symmetry group of a cube acts transitively and faithfully on its twelve edges. For vertices $(\pm 1,\pm 1,\pm 1)$, its signed permutation matrices act transitively on the edges. The stabilizer subgroup of $\{(1,1,z):-1\le z\le 1\}$ consists of independently swapping $x,y$ and reversing $z$, and is a Klein four-group. The orbit-stabilizer theorem gives order $12\cdot 4=48$. Fixing every edge fixes every vertex, as each vertex is the unique intersection of its three incident edges, so the group action is faithful. The resulting subgroup of $S_{12}$ has index $12!/48$ and is not normal: central inversion has cycle type $2^6$, whose conjugacy class in $S_{12}$ has $12!/(2^{6}6!)=10395>48$ elements.

This degree-four rational map obeys $R(iz)=1/R(z)$ and $R((iz+1)/(1-iz))=e^{2\pi i/3}R(z)$. These pair the generators of the rotational symmetry group of a cube with target rotations, giving combined spatial-isospin symmetry in the Skyrme model. Its spatial half-turn kernel is a Klein four-group. The map gives a cubic four-Skyrmion approximation; preserving only its Wronskian zero set does not guarantee the same rotational symmetry.

For a depressed quartic $x^4+p{x}^2+qx+r$ with roots ${\alpha }_1,\dots ,{\alpha }_4$, the three pair-partition quantities ${\alpha }_1{\alpha }_2+{\alpha }_3{\alpha }_4$ and its two analogues are the roots ofThe Galois action on these roots is the quotient of its action on the four quartic roots by the Klein four-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: $t\mathrm{mod}mn\mapsto (t\mathrm{mod}m,t\mathrm{mod}n)$.

Let the Klein four-group have generators $X,Y$ and $Z=XY$, with $X^2=Y^2=I$ and $XY=YX$. For every real $p$, the matricesdefine a group representation. Indeed, $D(Y{)}^2=I_2$, and $D(X)$ commutes with every matrix, so the group relations hold.

Set $S=\left(\begin{array}{cc}1& -p/2\\ 0& 1\end{array}\right)$. A direct change of basis givesConsequently all these representations are isomorphic representations, with character values $(2,-2,0,0)$ on $(I,X,Y,Z)$. Each is a direct sum of the two one-dimensional representations whose values on $X$ are $-1$ and whose values on $Y$ are $+1$ and $-1$. A group representation need not be orthogonal; nonzero $p$ changes the basis rather than the represented group.

The permutation matrices of the three displayed double transpositions, together with $I_4$, form a faithful representation of the Klein four-group. They satisfy $P^2=Q^2=I_4$ and $PQ=QP$. Every nonidentity element fixes no basis vector, so this is its four-dimensional regular representation.

The constant vector $(1,1,1,1{)}^T$ spans a nonzero proper invariant subspace, proving reducibility. More completely, the columns ofform a common orthonormal basis of eigenvectors. Direct multiplication gives $S^{T}PS=\mathrm{diag}(1,1,-1,-1)$ and $S^{T}QS=\mathrm{diag}(1,-1,1,-1)$. Thus the representation splits into the four distinct real one-dimensional characters, each once. The basis change is a normalized Hadamard matrix.

A generator of a group is a single element $g$ for which $G=⟨g⟩$: every group element is an integer power of $g$. A group admitting such an element is a cyclic group. In additive notation, every element must be an integer multiple of $g$.

Distinguish a single generator of a group from a generating set of a group. The Klein four-group needs two elements to generate it, although it has only four elements; no single element visits all four. A cyclic group need not have a unique generator of a group: $\mathbb{Z}/6\mathbb{Z}$ has $1$ and $5$, while the additive integers have $1$ and $-1$.

On the nonzero real numbers, sign reversal $a$ and reciprocation $b$ are commuting involutions. Their function composition generates exactly $e(x)=x$, $a(x)=-x$, $b(x)=1/x$ and $ab(x)=-1/x$. The relations $a^2=b^2=e$ and $ab=ba$ identify the group with $C_2\times C_2$, the Klein four-group. Its subgroups are the trivial group, its three order-two cyclic subgroups and itself. Excluding zero from the domain is necessary for all four transformations to be defined and invertible.

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.

Permuting the three pairs of opposite faces gives a group homomorphism from the rotational symmetry group of a cube to $S_3$. Its kernel consists of the identity and the half-turns about the three coordinate axes: these are precisely the diagonal sign matrices with determinant one. Thus it is a normal subgroup of order four, a Klein four-group.

The degree-four rational map $F(z)=z^2+z^{-2}$ has branch values $-2,2,\infty$. Away from their inverse images, the transformationsare commuting deck involutions. They generate a Klein four-group acting transitively on each fibreConsequently the monodromy group is $V_4$, acting by the identity and the three double transpositions.

The order of a finite group is its cardinality, written $|G|$. The identity element is included: $|\mathbb{Z}/6\mathbb{Z}|=6$, and the trivial group has order one, rather than zero.

Keep this count distinct from the order of a group element. In the additive finite cyclic group $\mathbb{Z}/6\mathbb{Z}$, the element $1$ has order six, $2$ has order three, $3$ has order two, and $0$ has order one. Each element's order is the least positive number of repetitions needed to return to the identity element $0$. The order of an element of a finite cyclic group computes these periods directly.

The count also constrains subgroups: Lagrange's theorem gives $|G|=[G:H]|H|$ for a subgroup $H$ of a finite group $G$, where the index of a subgroup counts its cosets. For $H=\{0,2,4\}$ inside $\mathbb{Z}/6\mathbb{Z}$, the two cosets are $\{0,2,4\}$ and $\{1,3,5\}$, so $6=2\cdot 3$. Divisibility is a necessary constraint; it does not imply that every divisor occurs as the order of a group element, as the Klein four-group shows.

Define $G^{(0)}=G$ and $G^{(i+1)}=[G^{(i)},G^{(i)}]$. A group is a soluble group if this derived series reaches the identity after finitely many steps. Its derived length is the least $d$ with $G^{(d)}=1$.

Let $V=\{1,(12)(34),(13)(24),(14)(23)\}\le A_4$. This Klein four-group is a normal subgroup and Abelian; $A_4/V$ has order three, so $A_4^{\prime }\subseteq V$. The group $A_4$ is nonabelian, so its derived subgroup is nontrivial. Conjugation by a three-cycle permutes the three nonidentity elements of $V$ transitively. Since the derived subgroup is a normal subgroup, containing any one forces it to contain all three. Thus $A_4^{\prime }=V$, and $V^{\prime }=1$. Therefore $A_4$ has derived length two.

If $H\le G$, every commutator in $H$ is one in $G$, and induction gives $H^{(i)}\subseteq G^{(i)}$ for all $i$. Thus a subgroup of a solvable group is solvable, with no larger derived length.

The sign map gives $S_4^{\prime }\subseteq A_4$. The commutator $[(12),(23)]$ is a three-cycle. Normality of $S_4^{\prime }$ supplies every conjugate three-cycle, and three-cycles generate $A_4$. Hence $S_4^{\prime }=A_4$, so the complete derived series isEvery inclusion is strict, proving solvability and derived length three.