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The intersection $H\cap K$ contains the identity element. If $a,b\in H\cap K$, then $a{b}^{-1}$ belongs to both $H$ and $K$, so it belongs to their intersection. The subgroup test proves that $H\cap K$ is a subgroup of $G$, with the inherited operation. Explicitly, closure and inverses come from this test, and associativity is inherited from $G$.

Label the vertices of the equilateral triangle $1,2,3$ counterclockwise. For the dihedral group of order six, let $r=(123)$ be the counterclockwise rotation by $120^{\circ }$, and let $s=(23)$ be reflection in the line through vertex 1 and the opposite side's midpoint. ChooseThen $B$ rotates by $240^{\circ }$. The three Euclidean reflections $C,D,E$ fix vertices $1,3,2$ respectively. Products act on a vertex from right to left. The relations $r^3=s^2=e$ and $sr=r^{-1}s$ givewhere exponents of $r$ are reduced modulo three and those of $s$ modulo two. This constructs the entire Cayley table:For example $AC=D$ whereas $CA=E$, so the group is nonabelian.

A subgroup of order two is generated by its unique nonidentity element, which must have order two. The rotations $A,B$ have order three, and each reflection has order two. Thus there are exactlyNone is a normal subgroup. Conjugation by $A$ sends the reflection $r^{j}s$ to $r^{j+2}s$, cycling the three reflections. For example $AC{A}^{-1}=E\notin ⟨C⟩$. Each of the other two subgroups fails the same conjugation test.

Finally, take two distinct such subgroups. For instance their union is $⟨C⟩\cup ⟨D⟩=\{I,C,D\}$, but $CD=B$ lies outside it. Thus it is not closed under the inherited group operation and is not a group. The distinctness is necessary: the union of a subgroup with itself is that subgroup.

The group order $|G|$ is the number of elements of the finite group. A normal subgroup $H\le G$ satisfies $gH{g}^{-1}=H$ for every $g\in G$, equivalently its left and right cosets agree.

For the square, let $R$ be counterclockwise rotation by $\pi /2$, and $s$ reflection in the horizontal axis. The dihedral group isThe first four elements rotate by $0,90,180,270$ degrees. The other four are Euclidean reflections in the horizontal axis, the line $y=x$, the vertical axis and the line $y=-x$, respectively, for a square centred at the origin with sides parallel to the axes.

The eight nontrivial proper subgroups areTo check completeness, Lagrange's theorem allows proper orders one, two and four. The order-two subgroups are generated by the five involutions displayed. An order-four subgroup containing $R$ or $R^3$ is the rotation subgroup. Otherwise it must contain reflections; their product is a rotation, which must be $R^2$, so its reflections differ by that half-turn, giving the two listed Klein subgroups. The trivial subgroup $\{I\}$ is an additional proper subgroup. Thus the PDF's count of eight uses “proper” to exclude the trivial subgroup; with the usual inclusive definition there are nine proper subgroups.

A normal subgroup of order two is central: conjugation fixes its identity and must also fix its only other member. The half-turn $R^2$ commutes with every square symmetry, whereas the reflection subgroups are exchanged by conjugation. Hence the unique order-two normal subgroup isFor the general polygon, write $D_n=\{R^j,R^{j}s:0\le j<n\}$ with $R^n=s^2=I$ and $sRs=R^{-1}$. A rotation $R^j$ commutes with $s$ exactly when $R^j=R^{-j}$, or $2j\equiv 0(\mathrm{mod}n)$. A reflection cannot commute with $R$ when $n\ge 3$, becauseTherefore the center of a group here is $\{I\}$ for odd $n$, and $\{I,R^{n/2}\}$ for even $n$. The order-two normal subgroups correspond exactly to nonidentity central involutions, provingThis agrees with the involutions in an even dihedral group and uses the convention $|D_n|=2n$ throughout.

A subgroup $N\le G$ is a normal subgroup when conjugation by every group element preserves it:Equivalently, $gN=Ng$ for every $g$; normality permits the cosets to inherit a quotient-group multiplication.

A group homomorphism is a map preserving multiplication:It consequently sends identity to identity and inverses to inverses, since $\varphi (e)=\varphi (e{)}^2$ and $\varphi (g)\varphi (g^{-1})=\varphi (e)$.

The conjugacy class of $g$ is $\{hg{h}^{-1}:h\in G\}$. Rotations commute with one another, while conjugation by a reflection changes $R^j$ to $R^{-j}$. Thus the rotation classes are $\{I\}$, $\{R,R^4\}$ and $\{R^2,R^3\}$.

For reflections,Because multiplication by two permutes the residues modulo five, these conjugates exhaust all five reflections. Consequently the complete class partition isTheir sizes $1+2+2+5=10$ account for every element, as in the conjugacy classes of a finite dihedral group.

A normal subgroup is a union of complete conjugacy classes. The trivial subgroup is normal, and the rotation subgroup is normal because conjugation preserves it; equivalently it has index two. None of the five order-two subgroups is normal, because it contains just one member of the five-element reflection class. Thus the proper normal subgroups are exactly

The conjugacy class of $g$ isIf $N$ is a normal subgroup and $g\in N$, then normality gives $hg{h}^{-1}\in N$ for every $h$. Thus the whole conjugacy class of each member of $N$ is contained in $N$, andConjugacy classes partition the group, so this is a union of complete disjoint classes, never selected fractions of classes.

The kernel is a normal subgroup, so it contains the identity class and is a union of the classes listed in part (c). Its candidate orders are therefore $1$, $1+3=4$, $1+2=3$ and $1+3+2=6$. Lagrange's theorem excludes four, since it does not divide six. The remaining possibilities areAll three are normal: the first and last trivially, while $A_3$ is the kernel of the map giving the sign of a permutation, or the index-two subgroup generated by a 3-cycle.

Because $\Phi$ is onto, the first isomorphism theorem gives $G^{\prime }\cong S_3/K$. Explicitly, the map $gK\mapsto \Phi (g)$ is well defined, multiplicative and surjective; equality of its values means $h^{-1}g\in K$, so it is also injective. ThusEach is realized by the identity map, the sign map and the trivial map, respectively. This completes the classification of normal subgroups and quotients of the symmetric group on three letters.

The quotient group $G/N$ consists of the cosets $gN$, with multiplicationFor well-definedness, replacing representatives by $g{n}_1$ and $h{n}_2$ changes their product to $gh(h^{-1}{n}_{1}h)n_2$. Since $N$ is a normal subgroup, the final two factors lie in $N$, so the product coset does not change.

Associativity follows from that in $G$; the identity element is the coset $N$, and $(gN{)}^{-1}=g^{-1}N$. Hence these cosets form a group. By Lagrange's theorem, every subgroup $H\le G$ has order dividing $|G|$, and its subgroup index is $[G:H]=|G|/|H|$. In particularNormality is needed for the quotient multiplication, whereas Lagrange’s counting formula holds for every subgroup.

Write $H=\{e,h\}$ with $h\ne e$. For any $g\in G$, the normal subgroup property implies $gh{g}^{-1}\in H$. This conjugate cannot be $e$, since multiplication by $g^{-1}$ and $g$ would then give $h=e$. There is only one remaining choice, soThus $h$ commutes with every element of $G$; the identity does too. HenceThis proves that a normal subgroup of order two is central. The proof does not require $G$ to be finite: conjugation permutes $H$ and fixes its identity, so it must fix its only other element.

A group homomorphism $\varphi :H\to G$ is a map preserving multiplication:It necessarily sends $e_H$ to $e_G$ and sends inverses to inverses, by applying this relation to $e_H^2$ and $h{h}^{-1}$. Its kernel of a group homomorphism isThe kernel is a normal subgroup of the domain: if $k\in \ker \varphi$ and $h\in H$, then $\varphi (hk{h}^{-1})=\varphi (h)e_G\varphi (h{)}^{-1}=e_G$. The kernel’s identity condition refers to the identity in the target group.

A member of $H$ is $A(t,b,t)$. The multiplication law from part (a) becomesIt stays in $H$ and is unchanged when the two factors are exchanged, since $ts=st$. The identity lies in $H$, and $A(t,b,t{)}^{-1}=A(-t,t^2-b,-t)$ also lies in $H$. Thus $H$ is an Abelian subgroup, namely the Abelian plane subgroup of a finite Heisenberg group in this coordinate convention.

There are five choices each for $t,b$, soThis verifies Lagrange's theorem: the subgroup order divides the group order, and the number of distinct left cosets is five.

Explicitly, the map $\phi (A(a,b,c))=a-c$ is a surjective group homomorphism to the additive group of ${\mathbb{F}}_5$, by the product formula. Its kernel of a group homomorphism is $H$. Thus $H$ is even a normal subgroup; the five left cosets are the level sets $a-c=d$ for $d=0,1,2,3,4$, with representatives $A(d,0,0)$. Each contains 25 elements, providing a direct count as well.

The two diagonal generators are independent involutions, soIt is a Klein four-group. Write the swapping generator as $J=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Then $J^2=I$, so $H_2=\{I,J\}$ has order two. Moreover,Thus $J$ normalizes $H_1$. The setis a group: its products reduce usingand inverses reduce in the same way. It contains both generating subgroups, so minimality gives $G\subseteq K$; conversely every displayed product is in $G$, giving $G=K$.

There are four diagonal matrices $D$ and four off-diagonal matrices $DJ$, all distinct. ConsequentlyAlso $H_1\cap H_2=\{I\}$, since $J$ is not diagonal, and every element of $G$ has the form $D{J}^{\eta }\in H_1{H}_2$. This proves condition (I), hence the exact factorization of a group. Here $H_1$ is a normal subgroup and $G=H_1⋊H_2$ is a semidirect product. Its matrices are the two-dimensional signed permutation matrices, also realizing the dihedral group of the square.

The two one-dimensional representations are the trivial representation, ${\rho }_{+}(g)=1$, and the sign representation, ${\rho }_{-}(g)=\mathrm{sgn}(g)$. The first respects multiplication immediately. For the second, the sign of a permutation is the determinant of its permutation matrix, and multiplicativity of the determinant proves ${\rho }_{-}(gh)={\rho }_{-}(g){\rho }_{-}(h)$. They are inequivalent since a transposition has images $1$ and $-1$, and conjugation by a nonzero scalar cannot change a one-dimensional matrix.

Their group homomorphism kernels areBoth are subgroups, indeed normal subgroups, because they are kernels. In the second case $A_4$ is the alternating group of even permutations, with twelve elements.

The characters are their scalar values, since the trace of a one-dimensional matrix is its only entry. In the class order from part (b),The character inner product is the group average of ${\chi }_{+}\overline{{\chi }_{-}}$, soEach has squared norm $(1+6+3+8+6)/24=1$, in agreement with character orthogonality for distinct irreducible representations.

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.

The kernel of a group homomorphism found in part (b) is the translation subgroupIt is a subgroup because $T_c{T}_d=T_{c+d}$, $T_0=I$ and $T_c^{-1}=T_{-c}$. It is nontrivial since $T_1\ne I$, and proper since, for example, $M(2,0)\notin N$.

Every kernel is a normal subgroup: if $n\in \ker f$, then $f(gn{g}^{-1})=f(g)ef(g{)}^{-1}=e$. Applying the same argument with $g^{-1}$ gives equality of the conjugate subgroup and the kernel. Here one can also check invariance under conjugation directly:Multiplication by nonzero $a$ permutes all $c\in \mathbb{C}$, so $M(a,b)NM(a,b{)}^{-1}=N$. Thus $N$ is a nontrivial proper normal subgroup, isomorphic to the additive group $(\mathbb{C},+)$.

For a subgroup $N\le H$ and $h\in H$, a left coset is $hN=\{hn:n\in N\}$ and a right coset is $Nh=\{nh:n\in N\}$. A normal subgroup satisfies $hN{h}^{-1}=N$ for every $h\in H$, equivalently $hN=Nh$.

If $N$ is normal, the quotient group is the set $H/N$ of cosets, with multiplicationTo check independence of representatives, replace $h,k$ by $h{n}_1,k{n}_2$. Thenbecause normality puts $k^{-1}{n}_{1}k$ in $N$. Thus the operation is well defined. Associativity follows from that in $H$, the identity is $N$, and the inverse of $hN$ is $h^{-1}N$. This proves the group axioms for $H/N$ and explains the role of coset multiplication is well-defined exactly for normal subgroups.

The subgroup generated by $T,P$ is $H=\{I,T,P,-I\}$, a Klein four-group. It is not normal in $G^{\prime }$. Indeed,For $x\ne 0$ the off-diagonal entries are nonzero, whereas every member of $H$ is diagonal. Thus this conjugate of $T$ is outside $H$, givingThe normal subgroup in part (c) is the continuous boost subgroup; choosing one representative from each of its cosets does not make those representatives a normal subgroup.

Let $n=p\ge 3$ be prime. Any subgroup contained in $⟨R⟩\cong C_p$ is either $\{I\}$ or the whole rotation subgroup. Both are normal: the trivial subgroup always is, and the rotation subgroup has index two, or equivalently is preserved by $mR{m}^{-1}=R^{-1}$.

If a normal subgroup contains any reflection, it contains its entire conjugacy class. Part (b) shows that this is every reflection for odd $p$. Products of reflections include a generator of the rotations, since $(Rm)m=R$. Such a normal subgroup must therefore be the whole dihedral group.

Thus the complete list of proper normal subgroups, including the trivial subgroup, isThis also agrees with classification of subgroups of a dihedral group: the reflection subgroups of order two are proper but are not normal when $p$ is odd.

For a subgroup $H\le G$ and $g\in G$, the left and right cosets are $gH=\{gh:h\in H\}$ and $Hg=\{hg:h\in H\}$. A normal subgroup satisfies $gH{g}^{-1}=H$ for every $g$. Multiplying this set equality on the right by $g$ givesConversely this equality implies $gH{g}^{-1}=H$, so it is an equivalent definition of normality.

For a normal subgroup $H$, the quotient group consists of its cosets with multiplicationTo check that this is well-defined, replace $g,k$ by $g{h}_1,k{h}_2$ with $h_1,h_2\in H$. Their product issince normality puts $k^{-1}{h}_{1}k$ in $H$. Hence the product does not depend on representatives. Associativity follows from that in $G$, the identity element is $H=eH$, and $(gH{)}^{-1}=g^{-1}H$. These verify the group axioms.

Lagrange's theorem counts the equal-sized cosets:Finally, if $|G|=p$ is prime, choose $g\ne e$. Its cyclic subgroup has order greater than one dividing $p$, hence equals $G$. All elements are powers of $g$, so $g^j{g}^k=g^{j+k}=g^k{g}^j$. Thus

LetThe product of any two distinct nonidentity elements is the third, and each squares to $e$, so $K$ is a Klein four-group. Conjugating a double transposition simply relabels its letters, keeping it in the set. Hence $K$ is a normal subgroup of $A_4$.

Choose $r=(123)$. The cosets $K,rK,r^{2}K$ are distinct and cover the twelve elements. With $\zeta =e^{2\pi i/3}$, defineThe coset decomposition makes this well-defined. Normality givesso $\Phi$ preserves products. It is onto $\{1,\zeta ,{\zeta }^2\}\cong C_3$, andThis is the tetrahedral quotient onto the cyclic group of order three. In terms of (a), it sends $C_0\cup C_2$ to $1$, $C_{+}$ to $\zeta$, and $C_{-}$ to ${\zeta }^2$. The normality check above also follows abstractly because a group homomorphism kernel is normal.