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With $\gamma (v)=(1-v^2/c^2{)}^{-1/2}$,Adding and subtracting $c{t}^{\prime }$ givesThe product of the two multipliers is one, so $x_{+}^{\prime }{x}_{-}^{\prime }=x_{+}{x}_{-}$, which is $x^{\prime 2}-c^2{t}^{\prime 2}=x^2-c^2{t}^2$.

Successive transformations multiply the $x_{+}$ factors. Equatingand solving gives the relativistic velocity-addition law

During a short time $dt$, the rocket loses mass $-dm>0$. Conservation of upward momentum, including gravity's impulse and exhaust velocity $v-U$, gives to first orderwhich is the stated rocket equation.

Here $dm/dt=-\alpha$ and $U=U_0{m}_0/m$, soLift-off from rest requires positive initial acceleration,With $v(0)=0$, integration givesThe dimensions are $[m_0]=[m]=M$, $[\alpha ]=M/T$, $[U_0]=[U]=[v]=L/T$, $[g]=L/T^2$, and $[t]=T$. Both displayed terms in $v$ therefore have dimension $L/T$.

The Chinese remainder theorem says that for pairwise coprime positive integers $m_i$, the mapis a bijection. For two moduli, choose $u,v$ with $um+vn=1$; thenhas residues $a$ modulo $m$ and $b$ modulo $n$. Uniqueness follows because the difference of two solutions is divisible by both coprime moduli, hence by their product. Induction proves the general case.

The two given congruences are compatible modulo $\gcd (6,8)=2$, and checking modulo $\mathrm{lcm}(6,8)=24$ gives

Write $d=2^r{3}^{s}m$ with $(m,6)=1$. Use the Chinese remainder theorem to chooseThen $4a^2+9b^2\equiv 1$ modulo each of the pairwise coprime factors $2^r$, $3^s$, and $m$, hence modulo $d$.

Assume Newton's second law, $F_{ij}=-F_{ji}$, and that each internal pair force is central, so $F_{ij}$ is parallel to $r_i-r_j$. Summingcancels internal pairs and gives $dP/dt=F:={\sum }_i{F}_i$. Taking moments about fixed $a$ cancels the internal torques pairwise and gives

The angular momentum about the centre of mass result follows similarly: differentiating ${\sum }_i(r_i-R)\times m_i({\dot{r}}_i-\dot{R})$ introduces no extra term because both total relative position weighted by mass and total relative momentum vanish. Thus its derivative is the external torque about $R$.

Finally, if every mass is $m$ and $F_i=-k{\dot{r}}_i$, then about a fixed pointTherefore

Using the product rule for directional derivatives,Thus $f=|u{|}^2/2-P$ and $g=-h$. If $\nabla \cdot u=0$ and $w=\nabla \times u$, then $\nabla \cdot w=0$. Integrating the two directional derivatives as divergences gives only boundary fluxes, which vanish because $u\cdot n=w\cdot n=0$. This is helicity conservation by tangent boundary conditions, so $dH/dt=0$.

For the stated field, direct use of the cylindrical curl gives $u\cdot w=-2a{\rho }^4{\sin }^{2}z$. Therefore

The first isomorphism theorem says that for a homomorphism $\varphi :G\to H$,The map is well-defined because equal cosets differ by a kernel element; it is a surjective homomorphism onto the image, and injectivity follows because its kernel is the identity coset.

Here $G{L}_n(\mathbb{R})$ consists of invertible real $n\times n$ matrices and $S{L}_n(\mathbb{R})$ consists of those with determinant one. The latter is the kernel of the surjective determinant map to ${\mathbb{R}}^{\ast }$, so it is normal and

The given integral matrices form a group because products and inverses remain integral. The inverse formula shows that an integral inverse exists exactly when the determinant divides every cofactor; taking determinants shows more directly that $\det A\det A^{-1}=1$, hence $\det A=\pm 1$, and the adjugate formula proves the converse. The matrices $\left(\begin{array}{cc}1& m\\ 0& 1\end{array}\right)$ show infinitude.

Reduction modulo two is a homomorphism $G\to G{L}_2({\mathbb{F}}_2)$. Its kernel is exactly $H$, so $H$ is normal; its index is finite because the target has only six elements.

The map $H\times K\to HK$, $(h,k)\mapsto hk$, has fibers of size $|H\cap K|$: two pairs have the same product exactly when their quotient is represented by an element of the intersection. Hence

The standard intersection-index inequality gives $[G:H\cap K]\le [G:H][G:K]=ab$. This index is divisible by both coprime numbers $a$ and $b$, so it equals $ab$. The displayed formula then gives $|HK|=|G|$, hence $HK=G$.

The conjugacy class and centralizer areand orbit--stabilizer gives $|{\mathrm{Conj}}_G(x)|=[G:C_G(x)]$. Thus the coprimality hypothesis gives $C_G(x)C_G(y)=G$.

One inclusion in the desired equality follows by conjugating $xy$. Conversely take $g^{-1}xg{h}^{-1}yh$. Write $h{g}^{-1}=ab$ with $a\in C_G(y)$ and $b\in C_G(x)$. The hint then rewrites the product aswhich is conjugate to $xy$. Therefore

Define $(xN)(yN)=xyN$. If representatives are changed to $x{n}_1,y{n}_2$, normality moves $n_1$ past $y$ and leaves the product in $xyN$, so multiplication is well-defined.

Let $G^{\prime }$ be the set of finite products of commutators. It is closed under products, and $[x,y{]}^{-1}=[y,x]$, so it is a subgroup. Moreoverso conjugation preserves products of commutators and $G^{\prime }◃G$. In $G/G^{\prime }$, one has $xy{G}^{\prime }=yx{G}^{\prime }$ because $(yx{)}^{-1}xy=[x,y]$, hence the quotient is abelian. Conversely, if $G/N$ is abelian, every commutator maps to the identity, so $G^{\prime }\le N$. This is the universal property of the commutator subgroup.