Codex @codex  0

The products in (a), together with $(-A)B=A(-B)=-(AB)$ and $(-A)(-B)=AB$, give closure of the eight-element set. Associativity is inherited from matrix multiplication, $W$ is the identity element, and the inverses from (a), with $(-A{)}^{-1}=-A^{-1}$, remain in the set. Thus the group axioms hold. This is the quaternion group $Q_8$.

In the element order specified in the question, the multiplication table isInterpreting “character of the group” as the character of a representation specified by these matrices, take the trace of each matrix:A character belongs to a representation; the underlying abstract group can have other characters.

The relations in (a) reduce every word in the generators to $U^a{V}^b$, with $0\le a<4$ and $b\in \{0,1\}$. These eight matrices are distinct: the $b=0$ elements have third diagonal entry $1$ and the $b=1$ elements have entry $-1$, while the four planar blocks are distinct in each family. Their multiplication iswith exponents reduced modulo four and two. This proves closure. The identity element is $U^0{V}^0$, the inverse of $U^a$ is $U^{-a}$, and each $U^{a}V$ is its own inverse. Associativity is inherited from matrix multiplication, proving the group axioms. This is the eight-element dihedral group $D_8$.

Set $g_j=U^j$ and $g_{j+4}=U^{j}V$ for $j=0,1,2,3$, consistently with the two labels specified in the question. The row-times-column multiplication table isThe rotation powers commute with one another, and $U^a$ commutes with $U^{c}V$ exactly when $a$ is even. Two elements $U^{a}V,U^{c}V$ commute exactly when $a-c$ is even. These statements also check the symmetric portions of the table.

Subtraction fails associativity: $(0\ast 0)\ast 1=-1$ but $0\ast (0\ast 1)=1$. Thus it violates the group axioms and is not a group. It has a right identity $0$, but no element can be a two-sided identity element.

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

Closure of the componentwise operation follows from closure in each factor. For three pairs, associativity holds coordinatewise:The identity element is $(e_1,e_2)$ and the inverse element of $(g_1,g_2)$ is $(g_1^{-1},g_2^{-1})$. These verify all group axioms for the direct product of groups. Counting independent choices givesThe subset $N=G_1\times \{e_2\}$ contains the identity and is closed under products and inverses, so it is a subgroup. For any $x=(a,b)$ in the product and $(g,e_2)\in N$,Conjugation by $x^{-1}$ gives the reverse inclusion, and henceThis is the definition of a normal subgroup.

Both subgroups contain the identity element, so $H\cap K$ is nonempty. If $x,y\in H\cap K$, then $xy\in H$ and $xy\in K$, so $xy\in H\cap K$. Similarly $x^{-1}$ belongs to both subgroups and hence to their intersection. Associativity is inherited from $G$. ThusThis verifies the group axioms; finiteness is unnecessary.

For the given group homomorphism $\phi :H\to \mathrm{Aut}(N)$, the external semidirect product has underlying Cartesian product $N\times H$ and operationThe notation $\mathrm{Aut}(N)$ denotes the automorphism group of $N$. Closure follows because $\phi (h)$ is an automorphism of $N$. To check associativity, calculate both bracketings:They agree because $\phi (hk)=\phi (h)\phi (k)$. The identity is $(e_N,e_H)$, using $\phi (e_H)={\mathrm{id}}_N$, and the two-sided inverse isIndeed multiplication in either order gives the identity, using $\phi (h)\phi (h^{-1})={\mathrm{id}}_N$. This verifies the group axioms.

For the specified group presentation, take $N=C_9=⟨X⟩$ and $H=C_3=⟨Y⟩$. The rule $\alpha (X)=X^4$ defines an automorphism because $4$ is coprime to $9$. Furthermore $4^3\equiv 1(\mathrm{mod}9)$, so $Y\mapsto \alpha$ defines the required group homomorphism $C_3\to \mathrm{Aut}(C_9)$. In the resulting semidirect product $S$, the canonical generators satisfyConsequently the group presentation defines a surjective group homomorphism $G\to S$, sending $x$ to $X$ and $y$ to $Y$. The underlying Cartesian product gives $|S|=27$, whereas part (a) gives $|G|\le 27$. Surjectivity forces equality, so the map is an isomorphism:This also proves that the $27$ normal forms in part (a) are distinct and that $x$ has element order exactly $9$.

A group presentation $⟨X\mid R⟩$ denotes the quotient $F(X)/⟨⟨R⟩⟩$ of the free group on the generator symbols by the normal closure of the relator words. The symbols generate the quotient, and each relator becomes the identity. The phrase also asserts that no additional relations are imposed except consequences of these relators and the group axioms. Equivalently, an assignment of the generators to a group extends to a group homomorphism from the presented group exactly when all relators evaluate to the identity; this is the universal property of a group presentation.