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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.

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.

The substitution $F(a)=e^a$, with $F(-\infty )=0$, is a bijection onto $[0,\infty )$ and obeysThus the log-sum-exp function transfers nonnegative addition to the given set. It is closed, associative and commutative, with identity element $-\infty$. But a finite $a$ has $e^a>0$, and adding a nonnegative number cannot give zero, so no finite $a$ has an inverse element. Therefore it is a commutative monoid but not a group.

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.

The sum of two odd integers plus one is odd, so the operation is closed. Both associations of a triple give $a+b+c+2$, proving associativity. Its identity element is $-1$, and the inverse element of $a$ is $-a-2$, also odd, since $a+(-a-2)+1=-1$. Thus it is an abelian group.

More explicitly, shifted addition on odd integers is ordinary integer addition in different coordinates:This bijection is a group isomorphism to $(\mathbb{Z},+)$.

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

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.

The group order is its cardinality, $\boxed{|G|=\#\{g:g\in G\}}$. The order of a group element $g$ iswhere $e$ is the identity element. In a finite group such an $n$ always exists, since two nonnegative powers of $g$ must coincide.

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.

Use group axioms from two-sided local inverse identities. Choose any $a$ in the nonempty semigroup and put $e_R=a_L^{-1}a$, $e_L=a{a}_R^{-1}$. The given identities say $b{e}_R=b$ and $e_{L}b=b$ for every $b$. Therefore $e_L{e}_R=e_L$ by the right-identity property, and $e_L{e}_R=e_R$ by the left-identity property. Thus $e_L=e_R=e$ is a two-sided identity element.

For an arbitrary $a$, set $b=a_L^{-1}$ and $c=a_R^{-1}$. Inserting $e$ into the original identities gives $ba=e$ and $ac=e$. By associativity,Thus every element has a two-sided inverse element. Closure and associativity were assumed, so all the group axioms hold. No cancellation or identity was used before it had been established.

A left-primitive ideal is a two-sided ideal of the formfor a nonzero simple module $V$ on the left. Equivalently, the quotient ring $R/P$ has a faithful module that is a simple module, and is therefore a primitive ring.

To prove that $P$ is a prime ideal of a noncommutative ring, let $A,B$ be two-sided ideals with $AB\subseteq P$. Since $B$ is two-sided, $BV$ is a submodule of $V$: for $r\in R$, $b\in B$ and $v\in V$, one has $r(bv)=(rb)v\in BV$. If $B⊈P$, then $BV\ne 0$, so simplicity gives $BV=V$. Consequentlyand $A\subseteq P$. Thus $AB\subseteq P$ implies $A\subseteq P$ or $B\subseteq P$. Also $P\ne R$, since $1$ acts as the identity element on the nonzero module $V$. These are precisely the required prime ideal of a noncommutative ring conditions.

A left group action of $G$ on a set $S$ is a map $G\times S\to S$, written $(g,s)\mapsto g\cdot s$, satisfyingfor every $s\in S$ and $g,h\in G$, where $e$ is the identity element of $G$. In particular each $g$ acts by a bijection, whose inverse is the action of $g^{-1}$.

The identity element satisfies $e\ast x=ex{e}^{-1}=x$. For $g,h\in G$,Both action axioms hold, so conjugation is a group action on $G$.

Conjugacy is an equivalence relation: the identity element gives reflexivity, conjugation by $g^{-1}$ reverses conjugation by $g$, and composing conjugations gives transitivity. Its equivalence classes therefore partition the finite set $G$. Counting this disjoint union givesSeparating singleton classes from the others gives the class equation $|G|=|Z(G)|+{\sum }_{|C_i|>1}|C_i|$.

Write $|G|=p^n$, where $n\ge 1$. By part (c), each conjugacy-class size divides $p^n$ and hence is a power of $p$. Every nonsingleton class consequently has size divisible by $p$. The class equation impliesThe identity element lies in the centre, so the centre is nonempty. Its size is a positive multiple of $p$, and thereforeThis proves that a nontrivial finite p-group has a nontrivial centre.

For a continuously differentiable diffeomorphism $\varphi :J\to I$ between intervals, the integration by substitution statesIt holds as an identity element of extended integrals for nonnegative measurable $h$. For signed or complex-valued measurable $h$, integrability on one side is equivalent to integrability after substitution, and the identity element then holds. The absolute derivative accounts for a decreasing substitution; on oriented finite intervals the equivalent formula uses ${\varphi }^{\prime }$ and oriented endpoints.

Put $C={\int }_0^{\infty }(1+x^4{)}^{-1}dx=\pi /(2\sqrt{2})$. The map $x=y^{-1/4}$ is a decreasing diffeomorphism of the positive half-line onto itself andApplying the nonnegative version before assuming finiteness proves both existence and value:Tonelli's theorem permits either iterated integral of a nonnegative measurable function on a product of sigma-finite measure spaces and identifies both with its extended product integral. Fubini's theorem states that for an absolutely integrable measurable function on such a product, almost every section is integrable, the section-integral functions are integrable, and the two iterated integrals equal the product integral.

The given two-variable function is continuous and positive on the positive quadrant. For fixed $y>0$, substituting $t=x{y}^{1/4}$ givesTonelli now givesThis proves absolute integrability, so Fubini is applicable.

For $x\ne 1$, integrate the displayed partial fraction identity element in $y$ up to $R$:At $x=1$ the section integral is ${\int }_0^{\infty }(1+y{)}^{-2}dy=1$, agreeing with the continuous limit of this expression. Therefore Fubini identifies the remaining integral and givesThe apparent singularity at one is removable, with limiting integrand value $1/4$. Positivity and the finite product integral already establish convergence at both ends.

The inverse derivative formula, evaluated at $G(0,0)=(0,0)$, givesMultiplying the two displayed matrices in either order gives the identity element, verifying the total derivative explicitly.

Orient the body boundary counterclockwise and write its tangent displacement as $(dx,dy)$. The outward normal from the body has displacement-weighted components $(dy,-dx)$. Pressure therefore exerts the force components $F_x=-\oint pdy$ and $F_y=\oint pdx$, orFor steady irrotational incompressible flow at constant density, Bernoulli gives $p=p_0-\rho |w^{\prime }(z){|}^2/2$, with any constant far-field kinetic term absorbed into $p_0$. Its constant contribution integrates to zero. The boundary is a streamline, so $\mathrm{Im}w$ is constant there and $w^{\prime }(z)dz$ is real. Thus $w^{\prime }dz=\overline{w^{\prime }}d\bar{z}$, giving $|w^{\prime }{|}^{2}d\bar{z}=(w^{\prime }{)}^{2}dz$. This identity element remains valid at a stagnation point as both sides vanish. Substitution yields Blasius force theorem: