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A real two-dimensional group representation is a group homomorphism $D:G\to GL(2,\mathbb{R})$, so $D(AB)=D(A)D(B)$ and $D(I)=I$.

Substitute $\tilde{x}=\alpha x+\beta y$ and $\tilde{y}=\gamma x+\delta y$ into the quadratic expression. Its coefficients of $x^2$ and $y^2$ are ${\alpha }^2\hat{a}+{\gamma }^2\hat{b}$ and ${\beta }^2\hat{a}+{\delta }^2\hat{b}$. The cross term is zero for every matrix in this particular cyclic group, because each has one nonzero entry per row. ThusLet $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ generate the group. For $A=\pm I$, $M=I$; for $A=\pm J$, $M=S:=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Since $S^2=I$, each $M$ is its own inverse and $D(J^k)=S^k$. The group law is respected: $S^{k+l}=S^k{S}^l$, and $S^4=I$ makes it independent of the representative modulo four. This proves that $D=M^{-1}$ is a group representation.

This quarter-turn action on diagonal quadratic forms hasso it is not faithful. Quadratic forms cannot distinguish a simultaneous sign reversal of both coordinates. The invariant coefficient directions $(1,1)$ and $(1,-1)$ correspond to the fixed form $x^2+y^2$ and the form $x^2-y^2$ whose sign changes under a quarter-turn.

For a subgroup $H$ of a finite group $G$, distinct left cosets are disjoint and partition $G$. Each coset $gH$ has exactly $|H|$ elements because $h\mapsto gh$ is a bijection. Counting the partition proves Lagrange's theorem:A cyclic group is generated by a single element. In a group of order four, an element of order four generates the whole group, giving $C_4$. Otherwise all three nonidentity elements have order two. Choose two of them $a,b$; cancellation makes $ab=c$ the third. Since $c^{-1}=c$, $(ab{)}^{-1}=b^{-1}{a}^{-1}=ba=ab$. The remaining products then give the Klein four-group. This proves that there are precisely two groups of order four up to isomorphism, distinguished by whether an element of order four exists.

For the square, take $r$ to be counterclockwise rotation by $90^{\circ }$ and $s$ reflection in the horizontal axis. They satisfy $r^4=s^2=e$ and $sr=r^{-1}s$. The dihedral group here is $D_4$ of order eight, with elements $e,r,r^2,r^3,s,rs,r^{2}s,r^{3}s$. Both generators are necessary: one generator would give a cyclic, hence abelian, group, while $r$ and $s$ do not commute. In the following Cayley table, the row acts after the column; the formula $(r^a{s}^b)(r^c{s}^d)=r^{a+(-1{)}^{b}c}{s}^{b+d}$, with exponents reduced modulo four and two, computes every entry.The largest cyclic subgroup is $⟨r⟩=\{e,r,r^2,r^3\}$, of order four. Every reflection has order two and there is no element of order eight. The other subgroups of order four areBoth are Klein four-groups, generated by the commuting involutions $r^2$ and one reflection. For completeness, any order-four subgroup containing $r$ or $r^3$ equals $⟨r⟩$. Otherwise it contains a reflection $f$; the product of two distinct reflections is a nonidentity rotation, which must be $r^2$ if odd powers of $r$ are excluded. This forces the subgroup to be $\{e,r^2,f,r^{2}f\}$, giving exactly the two displayed choices. It is the square case of the classification of subgroups of a dihedral group.

The order of a finite group $G$ is its number of elements. Lagrange's theorem states that for a subgroup $K\le G$,so $|K|$ divides $|G|$. In particular the order of a group element divides $|G|$, because it is the order of its generated cyclic subgroup.

For $|G|=4$, if some element has order four it generates the whole group, which is therefore the cyclic group $C_4$. Otherwise every nonidentity element has order two. Choose two distinct ones $a,b$ and put $c=ab$. Cancellation shows $c\notin \{e,a,b\}$, so $c$ is the remaining element. Since $c^{-1}=c$,The other products follow: $ac=ca=b$, $bc=cb=a$, and $a^2=b^2=c^2=e$. Thus the second possibility is the Klein four-group. The two groups are not isomorphic because only $C_4$ has an element of order four. This proves the order-four case of the classification of groups of order p squared.

A group homomorphism $\Phi :G\to H$ obeys $\Phi (gh)=\Phi (g)\Phi (h)$. It maps identity to identity and inverses to inverses. Its group kernel $K=\{g:\Phi (g)=e_H\}$ contains $e_G$ and is closed under $a{b}^{-1}$, so the subgroup test makes it a subgroup. For $k\in K$ and $g\in G$,Thus $gK{g}^{-1}\subseteq K$; applying the same argument to $g^{-1}$ gives equality. Hence $K$ is a normal subgroup.

Define $\overline{\Phi }:G/K\to H$ by $\overline{\Phi }(gK)=\Phi (g)$. If $gK=hK$, then $h^{-1}g\in K$, giving $\Phi (g)=\Phi (h)$, so this is well-defined. Coset multiplication shows it is a group homomorphism. Its kernel is just the identity coset, since $\Phi (g)=e_H$ means $g\in K$. It is therefore injective, and the assumed surjectivity of $\Phi$ makes it surjective as well. This proves the first isomorphism theorem here:

In the quaternion group, the element $-1=i^2=j^2=k^2$ commutes with each of $i,j,k$, because an element commutes with its square. It then commutes with their negatives too, and with $\pm 1$. The set $N=\{1,-1\}$ is a subgroup because $(-1{)}^2=1$, and it is central, hence a normal subgroup.

The four quotient cosets are $N,iN,jN,kN$. They square to $N$, and $(iN)(jN)=kN$. Reversed products have the same coset: for example $(ij{)}^{-1}=k^{-1}=-k$, while $j^{-1}{i}^{-1}=(-j)(-i)=ji$, so $ji=-k$. The same calculation gives $kj=-i$ and $ik=-j$. After signs are identified, these are precisely the products in the Klein four-group. DefineThis preserves all products, is surjective, and has kernel exactly $N$. The first isomorphism theorem therefore givesThis is the quaternion group modulo its centre: the reverse-product relations also show that none of $\pm i,\pm j,\pm k$ is central, so $Z(Q_8)=N$.

The group order of a finite group is its number of elements, $|G|$. The order of a group element $g$ is the least positive integer $r$ with $g^r=e$. Such an $r$ exists in a finite group because two powers must coincide, and cancellation then gives a positive power equal to the identity element.

For the direct product of groups, the componentwise operation is closed because each factor is closed. Associativity holds in each coordinate:The identity element is $(e_1,e_2)$, and the inverse element of $(g,h)$ is $(g^{-1},h^{-1})$. Thus all group axioms hold. For finite factors, the multiplication principle gives

For ${\mathbb{Z}}_2\times {\mathbb{Z}}_2$, write $E=(e,e)$, $u=(w,e)$, $v=(e,w)$, $z=(w,w)$, with $w^2=e$. Its Cayley table, with the row element multiplying the column element, isThis is the Klein four-group: every nonidentity element has order two.

In the cyclic group ${\mathbb{Z}}_4=⟨a⟩$, the set $\{e,a^2\}$ is closed because $(a^2{)}^2=e$, and both elements have inverses in the set. It is therefore a proper subgroup of order two. By Lagrange's theorem, a proper subgroup has order one or two. An order-two subgroup must be generated by an element of order two. Neither $a$ nor $a^3$ has order two, whereas $a^2$ does. HenceThe PDF's claim needs “nontrivial”: $\{e,a^2\}$ is the only nontrivial proper subgroup, but $\{e\}$ is also a proper subgroup under the usual definition.

A group isomorphism preserves element orders: $\varphi (g{)}^r=e$ if and only if $g^r=e$, using the homomorphism property and injectivity. Since ${\mathbb{Z}}_4$ has elements of order four and the Klein four-group does not,

Write the finite cyclic group as $C_n=⟨g:g^n=e⟩$, the source's $Z_n$. It is Abelian, so $hg{h}^{-1}=g$ and every conjugacy class is a singleton: there are $n$ such classes.

Over the complex numbers, all its irreducible representations are one-dimensional. One direct proof is that the matrix representing $g$ has an eigenvector over $\mathbb{C}$; its line is preserved by every power of that matrix and hence by the whole group. Irreducibility forces the line to be the whole representation space. Its eigenvalue must be an $n$th root of unity.

The complex irreducible representations of a finite cyclic group are thereforeTheir generator values are distinct, so no two are equivalent. There are $n$ of them, and $n$ squared dimensions equal one, exhausting the sum of squares of irreducible representation dimensions: ${\sum }_j{d}_j^2=n=|G|$. Thus the list is complete.

For $n=3$, let $\omega =e^{2\pi i/3}$, so ${\omega }^3=1$ and $1+\omega +{\omega }^2=0$. The character table isThe complex-field convention is essential. The preceding part used real matrices, but that does not make the one-dimensional classification valid over $\mathbb{R}$: conjugate nonreal characters combine into the two-dimensional irreducible real representations of a finite cyclic group. The character table and dimension-sum result used here have their usual complex meaning.

A cyclic group of order $p$ contains an element of order two if and only if $2$ divides $p$. Necessity follows from Lagrange's theorem; for even $p$, the unique such element is $c=g_1^{p/2}$. HenceBecause $g_1$ has order $p>2$ and the representation is faithful, $D(g_1)$ cannot be a reflection, which has order two. It is therefore a rotation through a primitive angle $\vartheta =2\pi q/p$, with $\gcd (q,p)=1$. The condition on the greatest common divisor is exactly what makes its order $p$ rather than a proper divisor. Since $p$ is even, $q$ is odd, so $D(c)$ is the half-turn:The character is the trace:Choosing the primitive generator with $q=1$ gives the customary rotation representation. The general coprime value is included because faithfulness alone does not select that generator. These are the two-dimensional cases in the irreducible real representations of a finite cyclic group.

Direct matrix multiplication gives $P^2=P-I$ and hence $P^3=-I$. It follows thatand $P^6=I$. These six matrices are distinct; consequently no smaller positive power is the identity element. ThusProducts add exponents modulo six and $P^{-1}=P^5$. This verifies the cyclic group structure explicitly.

A group is a set $G$ equipped with a binary operation satisfying the group axioms: closure; associativity, $(gh)k=g(hk)$; an identity element $e$ with $eg=ge=g$; and an inverse element $g^{-1}$ for every $g$, with $g{g}^{-1}=g^{-1}g=e$.

A subgroup $H$ is a subset that is a group under the restricted operation. Equivalently, it is nonempty and contains $x{y}^{-1}$ whenever $x,y\in H$. A generating set of a group is a set whose elements and inverses can be multiplied in finite words to produce every group element.

The group order is its number of elements, $|G|$. The order of a group element $g$ is the least positive integer $r$ with $g^r=e$, or infinity if there is none. Lagrange's theorem states that for a finite group and subgroup $H$, $|H|$ divides $|G|$ and $|G|=[G:H]|H|$. In particular, every finite element order divides the group order.

A cyclic group is generated by one element, so every element is a power of that generator. A group isomorphism is a bijective group homomorphism: a bijection $\varphi :G\to G^{\prime }$ satisfying $\varphi (gh)=\varphi (g)\varphi (h)$ for all $g,h$.

A cyclic group is generated by one element: $G=⟨g⟩=\{g^n:n\in \mathbb{Z}\}$. The hypothesis needs the qualification no proper nontrivial subgroup: every nontrivial group has the proper trivial subgroup $\{e\}$, so the literal wording admits no examples.

Under that intended hypothesis, choose $g\ne e$. The subgroup $⟨g⟩$ is nontrivial and must therefore be all of $G$, proving cyclicity. If $g$ had infinite order, then $⟨g^2⟩$ would be nontrivial and proper, since $g$ is not an even power of an infinite-order generator. Thus $g$ has finite order $N>1$ and $|G|=N$.

If $N$ were composite, choose a divisor $d$ with $1<d<N$. Then $⟨g^d⟩$ would have order $N/d$, strictly between one and $N$, again contradicting the hypothesis. ConsequentlyConversely, Lagrange's theorem shows that a group of prime order has only itself and its trivial subgroup. This proves the characterization of groups with no proper nontrivial subgroups, even without assuming finiteness at the outset.

By (d), every group of group order three is a cyclic group. Writing its elements as $e,a,a^2$, every product is determined by $a^3=e$; the multiplication table isThe map $a^j\mapsto j(\mathrm{mod}3)$ is a bijective group homomorphism to addition modulo three. Thereforeso there is exactly one such group up to isomorphism.

A group homomorphism from the additive cyclic group ${\mathbb{Z}}_6$ is determined by the image $t$ of its generator $1$, with the consistency condition $6t=0$ in the target. Every element of either target satisfies this condition. ThereforeandThese formulas are well-defined because changing $n$ by six changes $tn$ by a target multiple of its modulus. They list respectively two and three maps.

Let $\alpha =p^{1/p}$ and let $\zeta$ be a primitive $p$th root of unity. The roots are $\alpha {\zeta }^j$, so $L=\mathbb{Q}(\alpha ,\zeta )$. Eisenstein gives $[\mathbb{Q}(\alpha ):\mathbb{Q}]=p$, while the prime cyclotomic extension has degree $[\mathbb{Q}(\zeta ):\mathbb{Q}]=p-1$. Both degrees divide $[L:\mathbb{Q}]$, and they are coprime. Conversely adjoining $\alpha$ to the cyclotomic field has degree at most $p$. These two bounds giveRestriction to $E=\mathbb{Q}(\zeta )$ gives a normal subgroup $N=\mathrm{Gal}(L/E)$ of order $p$, hence cyclic, with quotient $\mathrm{Gal}(E/\mathbb{Q})\cong C_{p-1}$. Restriction is onto because embeddings of the intermediate field extend in the finite Galois splitting field. Thus $1⊴N⊴G$ has abelian factors, proving $\boxed{G\text{ is solvable}}$. For $p=2$ the cyclotomic field is just $\mathbb{Q}$, and the same argument gives degree two and a cyclic group.

Let $a=\sqrt[n]{2}$ and $F=\mathbb{Q}(\xi )$. The roots of $X^n-2$ are $a{\xi }^j$, so $L=F(a)$. This splitting extension is separable in characteristic zero, hence Galois over $\mathbb{Q}$. The subextension $F/\mathbb{Q}$ is itself Galois by part (iii), so restriction gives a surjectionwith normal kernel $H=\mathrm{Gal}(L/F)$. Surjectivity follows from the extension theorem for embeddings and normality of $L/\mathbb{Q}$.

For $\sigma \in H$, write $\sigma (a)=a{\xi }^{j(\sigma )}$. This defines an injective homomorphism $H\to ⟨\xi ⟩$: injectivity holds because $L=F(a)$, and multiplicativity holds because $H$ fixes all roots of unity in $F$. A subgroup of a cyclic group is cyclic. The first isomorphism theorem therefore givesThere is no need to assume that $|H|=n$; the intersection with the cyclotomic field can reduce its order.

Let $\omega =(1+\sqrt{-3})/2$, whose minimal polynomial is $x^2-x+1$. For $p_2\equiv 1(\mathrm{mod}3)$, the cyclic group ${\mathbb{F}}_{p_2}^{\times }$ has a nontrivial cube root of unity $t$. Then $-t$ and $-t^2$ are distinct roots of $x^2-x+1$. By the Kummer-Dedekind theorem,Since the given ring of integers is a principal ideal domain, write $Q=(a+b\omega )$. Its norm isMultiplication by $4$ completes the square:Thus $x=2a+b$ and $y=b$ supply the required integers.

For a regular tetrahedron, rotations fixing a vertex rotate about the line from that vertex through the opposite face centre. Their angles are $0,2\pi /3,4\pi /3$, so the stabilizer is a cyclic group of order three. Rotations act transitively on the four vertices, giving $|G_1|=4\cdot 3=12$ by the orbit-stabilizer theorem.

For a cube, the corresponding axis is the body diagonal through the fixed vertex and its opposite. The three incident edges can be cyclically permuted by the same three angles, so again the stabilizer is cyclic of order three. The orbit has eight vertices. Hence

A cyclic group of order six has exactly one subgroup of each order $1,2,3,6$, and no others. Since it is abelian, these are also its four conjugacy classes of subgroups. Thus there is one transitive group action of each size $1,2,3,6$.

An action on six points is a disjoint union of its orbits. Its equivalence type is therefore determined by nonnegative integers satisfyingIf $n_6=1$ there is one possibility. If $n_6=0$, the possibilities $n_3=0,1,2$ allow respectively $4,2,1$ choices of $n_2$, with $n_1$ then fixed. HenceThis is counting finite group actions by transitive orbit types. Equivalently, a generator acts as a permutation whose cycles have lengths dividing six, and these are its eight possible cycle types on six points.

A cyclic group of order six has exactly one subgroup of each order $1,2,3,6$, and no others. Since it is abelian, these are also its four conjugacy classes of subgroups. Thus there is one transitive group action of each size $1,2,3,6$.

An action on six points is a disjoint union of its orbits. Its equivalence type is therefore determined by nonnegative integers satisfyingIf $n_6=1$ there is one possibility. If $n_6=0$, the possibilities $n_3=0,1,2$ allow respectively $4,2,1$ choices of $n_2$, with $n_1$ then fixed. HenceThis is counting finite group actions by transitive orbit types. Equivalently, a generator acts as a permutation whose cycles have lengths dividing six, and these are its eight possible cycle types on six points.

For a regular tetrahedron, rotations fixing a vertex rotate about the line from that vertex through the opposite face centre. Their angles are $0,2\pi /3,4\pi /3$, so the stabilizer is a cyclic group of order three. Rotations act transitively on the four vertices, giving $|G_1|=4\cdot 3=12$ by the orbit-stabilizer theorem.

For a cube, the corresponding axis is the body diagonal through the fixed vertex and its opposite. The three incident edges can be cyclically permuted by the same three angles, so again the stabilizer is cyclic of order three. The orbit has eight vertices. Hence

The cyclic group of order six has two different composition series:Their factor lists are $C_2,C_3$ and $C_3,C_2$ respectively; normality follows from commutativity.

The Jordan–Hölder theorem states that any two composition series of a group have the same length and the same multiset of composition factors up to isomorphism. Equivalently, the factors of one series can be reordered to match those of the other by group isomorphisms. The actual intermediate subgroups need not be the same.

Let $C_n=⟨g⟩$ and take $H=⟨g^p⟩$. Since $n=mp$, the order of $g^p$ is $m$, so $H\cong C_m$. All subgroups of a cyclic group are normal. The quotient is generated by $gH$ and has order $n/m=p$, givingA group of prime order has no nontrivial proper subgroup. By the subgroup correspondence theorem, no normal subgroup lies strictly between $H$ and $C_n$. Thus $H$ is a maximal proper normal subgroup.