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Every $D_2(g)$ is invertible because both inversion and the conjugate transpose preserve invertibility. For $g,h\in G$, the group homomorphism property of $D_1$ and reversal of multiplication under the adjoint giveAlso $D_2(e)=I$. Hence $D_2$ is a group representation. It is the inverse-adjoint representation of $D_1$; the inverse and adjoint each reverse product order, so together they preserve it.

Let $n=\mathrm{ord}(a)$, which is finite. A group homomorphism preserves powers and the identity, soIf $r=\mathrm{ord}(f(a))$, divide $n$ by $r$: $n=qr+s$ with $0\le s<r$. Then $e^{\prime }=f(a{)}^n=f(a{)}^s$, and minimality of $r$ forces $s=0$. ThusIf $f$ is a group isomorphism, apply the same argument to its inverse: $\mathrm{ord}(a)\le \mathrm{ord}(f(a))$. The two inequalities yield

The symmetric group $S_n$ consists of all bijections of $\{1,\dots ,n\}$, with composition as its operation; its group order is $n!$.

An $m$-dimensional complex group representation of $G$ is a group homomorphism $\rho :G\to GL(V)$, where $V$ is an $m$-dimensional complex vector space. After a choice of basis, each $\rho (g)$ is an invertible $m\times m$ complex matrix, with $\rho (gh)=\rho (g)\rho (h)$.

An irreducible representation has $V\ne 0$ and no invariant vector subspace other than $\{0\}$ and $V$. Two group representations $\rho$ on $V$ and $\sigma$ on $W$ are isomorphic representations if there exists an invertible linear map $S:V\to W$ such that $S\rho (g)=\sigma (g)S$ for every $g$. They are inequivalent if no such map exists; in equal-dimensional matrix form this says no fixed change of basis conjugates every $\rho (g)$ to $\sigma (g)$.

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 group homomorphism $f:G\to G^{\prime }$ satisfies $f(gh)=f(g)f(h)$ for all $g,h\in G$. It consequently sends the identity element to the identity and inverses to inverses. Its group homomorphism kernel isA normal subgroup $H$ of $G$ is a subgroup invariant under conjugation:Equivalently, $gH=Hg$ for every $g$, so left and right cosets coincide.

Write $M(a,b)=\left(\begin{array}{cc}a& b\\ 0& 1\end{array}\right)$. Direct matrix multiplication givesSince $a,c\ne 0$, the product is again in $G$. Associativity is inherited from matrix multiplication, the identity element is $M(1,0)$, andlies in $G$. These verify all group axioms. This is the Affine group of the complex line, acting by $z\mapsto az+b$.

For the diagonal subset, $M(a,0)M(c,0)=M(ac,0)$, it contains $M(1,0)$, and $M(a,0{)}^{-1}=M(a^{-1},0)$. Thus $G^{\prime }$ is a subgroup. Finally,so $f:G\to G^{\prime }$ is a surjective group homomorphism.

A rigid rotation of a regular tetrahedron permutes its four vertices, giving an injective group homomorphism from its rotation group to $S_4$: fixing every vertex forces the rotation to be the identity. Its seven nontrivial rotation axes are four lines through a vertex and the opposite face centre, and three lines through midpoints of opposite edges.

Each vertex-face axis admits rotations by $\pm 2\pi /3$, giving eight three-cycles on the vertices. Each opposite-edge axis admits a half-turn, giving three double transpositions. Along with the identity there are twelve rotations. All their induced permutations are even, so the injective image is a twelve-element subgroup of $A_4$, hence the whole group. ThereforeIts action on ${\mathbb{R}}^3$ supplies a three-dimensional representation. To prove irreducibility even after complexification, choose verticesThe group contains the half-turns $D_x=\mathrm{diag}(1,-1,-1)$ and their cyclic variants, as well as the coordinate cycle $(x,y,z)\mapsto (y,z,x)$. If a nonzero invariant subspace over the complex numbers contains a vector with nonzero $x$ component, applying $(I+D_x)/2$ shows it contains the $x$ axis. If another component is nonzero, use the corresponding projection instead. The coordinate cycle then supplies the other two axes. Every nonzero invariant subspace is therefore the full space, provingThis also gives an elementary proof without presupposing the character table.

Map each symmetry to the permutation it induces on the three vertices. This is a group homomorphism, because composing geometric motions composes their vertex permutations. In the labels chosen in (b), the map isAll six images are distinct and exhaust the symmetric group $S_3$. Equivalently, the kernel is trivial because a triangle symmetry fixing all three vertices is the identity. Thus the map is bijective and

A group homomorphism preserves multiplication:It automatically sends the identity element to the identity and sends inverse elements to inverses. A group isomorphism is a bijective group homomorphism. For finite groups this requires equal group orders, and, once those orders agree, either injectivity or surjectivity of the group homomorphism is sufficient. Equal orders alone do not suffice.

Here $a^2=-I$, $a^3=-a$ and $a^4=I$, so $K$ is a cyclic group of order four. Its multiplication table isWrite the specified image of $a$ as $A$. Direct matrix multiplication givesA group representation is a group homomorphism to a general linear group, so the images of all powers of $a$ are forced:All four images are invertible. Defining $\Phi (a^j)=A^j$ is well-defined modulo four because $A^4=I_2$, and $A^{j+\ell }=A^j{A}^{\ell }$ verifies every product in the table. These observations prove both existence and uniqueness of this representation with the prescribed image.

Distinguish the quaternion group generators $\mathbf{i},\mathbf{j},\mathbf{k}$ from the complex scalar $i$. Denote their proposed image matrices byDirect multiplication givesFor example $AB=\mathrm{diag}(i,-i)=C$ and $ABC=C^2=-I_2$. Sending $I$ to $I_2$ and each negative quaternion element to the negative of its positive image now reproduces every quaternion product. The matrices all have determinant one and are invertible, so this is a group representation in $GL(2,\mathbb{C})$.

The eight matrices $\pm I_2,\pm A,\pm B,\pm C$ are distinct. Therefore the group homomorphism is injective, and the representation is faithful; equivalently its kernel is $\{I\}$.

The character of a representation is the trace of its representing matrix. ConsequentlyThese are constant on the five conjugacy classes $\{I\}$, $\{-I\}$, $\{\pm \mathbf{i}\}$, $\{\pm \mathbf{j}\}$, $\{\pm \mathbf{k}\}$.

The complete list of subgroups isTo see completeness, Lagrange's theorem allows only orders $1,2,4,8$. The only element of order two is $-I$, which fixes the order-two subgroup. Every noncentral element has order four and generates the corresponding listed cyclic subgroup. Any order-four subgroup contains such an element, because there are only two central elements; it must equal the cyclic subgroup generated by that element. This accounts for all possibilities.

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.

In additive notation every element $(a,b)\in {\mathbb{Z}}_2\times {\mathbb{Z}}_4$ obeys $4(a,b)=(0,0)$. Its order of a group element therefore divides four, whereas $1\in {\mathbb{Z}}_8$ has order eight. A group isomorphism preserves element orders, soFor the second product, define the group homomorphismIt is well-defined because multiples of six are divisible by both two and three. Its kernel contains only zero: simultaneous divisibility by two and three implies divisibility by six. Thus $\Psi$ is injective, and equal finite group orders make it surjective. Explicitly the six images areThereforeFor general positive integers $n,m$, the order of a direct-product element is the least common multiple of its component orders. Hence every element order divides $\mathrm{lcm}(n,m)=nm/\gcd (n,m)$, while $(1,1)$ attains this order. It generates the whole product precisely when $n,m$ are coprime:This is the cyclicity of a product of two finite cyclic groups, with the explicit isomorphism furnished by the Chinese remainder theorem.

A group homomorphism $f:G\to H$ preserves the operation: $f(ab)=f(a)f(b)$ for all $a,b\in G$. A group isomorphism is a bijective group homomorphism; its inverse also preserves the operation. The kernel of a group homomorphism is $\ker f=\{g\in G:f(g)=e_H\}$, a normal subgroup of $G$.

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.

The symmetric group $S_3$ is generated by transpositions, each of order two. A group homomorphism to ${\mathbb{Z}}_3$ must map every transposition to zero, because the only target element annihilated by two is zero. Hence the map to ${\mathbb{Z}}_3$ is necessarily trivial.

For $S_3\to {\mathbb{Z}}_2$, the 3-cycles map to zero because their order divides three. All transpositions are conjugate group elements, and a map to an abelian group takes conjugate elements to the same value. That common value is either zero or one. Both choices work: they give the trivial map and the sign homomorphism, with an odd permutation sent to $1$ and an even permutation sent to $0$.

Let $G$ be the subgroup preserving the three-point set. Its action gives a group homomorphism $\Phi :G\to S_3$. If $\Phi (h)$ is the identity permutation, then $h$ fixes all three points, so part (i) forces $h$ to be the identity Möbius map. Thus the kernel is trivial and $\Phi$ is injective.

For any permutation $\sigma$, part (i) gives a normalization $g_{\sigma }$ sending the ordered triple $(z_{\sigma (1)},z_{\sigma (2)},z_{\sigma (3)})$ to $(0,1,\infty )$. If $g$ normalizes the original triple, then $h_{\sigma }=g_{\sigma }^{-1}\circ g$ sends $z_j$ to $z_{\sigma (j)}$. Hence every permutation is attained, so $\Phi$ is surjective.This is the setwise stabilizer; its six maps need not fix each point individually.

Given a group action, define $\rho (g)(j)=g\cdot j$. Each $\rho (g)$ is a permutation, since $\rho (g^{-1})$ is its inverse. The action law gives, for every $j$,Thus $\boxed{\rho :G\to S_n\text{ is a group homomorphism}}$.

Conversely, given a group homomorphism $\rho :G\to S_n$, put $g\cdot j=\rho (g)(j)$. Since a homomorphism sends the identity to the identity permutation, $e\cdot j=j$. AlsoThese are exactly the two group action axioms, so the construction defines the required action.

The three nonzero vectors in ${\mathbb{F}}_2^2$ are $e_1$, $e_2$ and $e_1+e_2$. An invertible linear map permutes this three-element set, so part (b) gives a group homomorphismIf a matrix acts trivially on all three vectors, it fixes both basis vectors $e_1,e_2$, so it is the identity matrix. Thus the kernel of a group homomorphism is trivial and $\rho$ is injective.

To count the domain, its first column may be any of the three nonzero vectors. The second column must be outside the two-element span of the first, leaving two choices. HenceAn injection between these finite sets of equal size is surjective, proving

Let $G$ act by left multiplication on its five left cosets of $H$. This gives a group homomorphism $\phi :G\to S_5$. Its kernel is the core of a subgroup, ${\bigcap }_{g\in G}gH{g}^{-1}$: an element fixes every coset precisely when all its conjugates lie in $H$. It is a normal subgroup contained in $H$, so the hypothesis forces the kernel to be trivial. Thus $G$ embeds in $S_5$.

The image has order $60$ and index two, hence is the kernel of a surjection $S_5\to C_2$. All transpositions are conjugate and generate $S_5$, so under any such surjection they must all have the same nonidentity image. This determines the map uniquely as the sign homomorphism. Its kernel is $A_5$, proving $G\cong A_5$.

To prove simplicity directly, the nonidentity conjugacy classes in $A_5$ have sizes $15,20,12,12$. A double transposition has centralizer of order four in $A_5$; a three-cycle has centralizer of order three; a five-cycle has centralizer of order five, yielding the two classes of size twelve. The class sizes together with the identity account for all sixty elements. A normal subgroup must be a union of these classes containing the identity, and its order must divide sixty. None of the proper nontrivial sums of these sizes with one is a divisor of sixty. Therefore $A_5$, and hence $G$, is simple.