The group $G$ is a nontrivial finite p-group, so the nontrivial center of a finite p-group gives $Z(G)\ne \{e\}$. If $Z(G)\ne G$, then $Z(G)$ is already a nontrivial proper normal subgroup, so $G$ is not a simple group. If $Z(G)=G$, then $G$ is abelian. By Cauchy's theorem for finite groups, it has a subgroup of order $p$; every subgroup of an abelian group is normal, and this subgroup is nontrivial and proper because $|G|=p^4$. Thus in every case

Put $ϵ=l/R$ and $a=1-\cos {\theta }_0$. ThenThe binomial series givessoSince $v_1^2=2gRaϵ$, a second binomial series expansion yieldsThus the constants in the stated Big O notation expansion are

Let $C$ be the centre of the Earth and let $M$ be its mass. Since $CW=R$ and $CS=R+l$, the law of cosines in triangle $CSB$ gives the ball's initial distance from $C$ asThe surface value of the Newtonian gravitational field satisfies $g=GM/R^2$. Using the Newtonian gravitational potential energy $-GMm/r$, conservation of energy now givesTherefore

The rod imposes a mechanical constraint, and its force is perpendicular to the wrecking ball's instantaneous velocity, so it does no work. In the constant parallel Newtonian gravitational field, the ball falls through the vertical distance $l-l\cos {\theta }_0$. Conservation of energy between $B$ and $W$ therefore givesand hence

The conclusion remains true, even though the scaling factor and hence the radius may now be arbitrary. At the point where its handle meets its circle, every lollipop contains a simple triod: take a short arc of the circle in each direction together with a short segment of the handle. Pairwise disjoint lollipops therefore contain pairwise disjoint simple triods. The Moore triod theorem says that every such family in the plane is countable, so

Fix a positive rational radius $r$. Distinct disjoint circles of radius $r$ have centers more than $2r$ apart, so their open interior disks are pairwise disjoint. Each such disk contains a point of the countable dense set ${\mathbb{Q}}^2$. Fix an enumeration of ${\mathbb{Q}}^2$ and assign to each disk the first rational point it contains. Disjointness makes this assignment injective, so there are only countably many circles of radius $r$.

The positive rational numbers are countable. Taking the countable union of countable sets over all rational radii proves that the whole collection is countable.

A set is countable if it is finite or admits an injection into the natural numbers; equivalently, its elements can be listed by a finite list or a sequence.

The product of two countable sets is countable: after choosing enumerations, its elements are indexed by pairs $(m,n)\in {\mathbb{N}}^2$, and these pairs can be listed diagonally. Thus an injection $X\to A\times B$ followed by an injection $A\times B\to \mathbb{N}$ proves that $X$ is countable.

The integers are listed as $0,1,-1,2,-2,\dots$, so $\mathbb{Z}$ is countable. Every rational number is represented by a pair $(p,q)\in \mathbb{Z}\times {\mathbb{N}}_{>0}$, so $\mathbb{Q}$ is countable as the image of a countable set.

Finally, if $A_0,A_1,\dots$ are countable, choose listings $A_n=\{a_{n,0},a_{n,1},\dots \}$, allowing repetitions for finite sets. The diagonal listing of the pairs $(n,m)$ lists every $a_{n,m}$, proving that a countable union of countable sets is countable.

DefineCertainly $f(B)\subseteq B$. If $y\in B$, then $y\in f(X)$ and has a unique predecessor $x$ because $f$ is injective. For every $n$, writing $y=f^{n+1}(z)$ and using uniqueness gives $x=f^n(z)$; hence $x\in B$ and $y\in f(B)$. Thus $f(B)=B$.

If $f(a)\in B$, its unique predecessor must lie in $B$, so $a\in B$; consequently $f(A)\subseteq A$. If a point belonged to every $f^n(A)$ for $n\ge 1$, it would belong to $B$ and also to $f(A)\subseteq A$, which is impossible. Therefore

The class of $(1,1)$ isso its size is Euler's totient functionThe unit group acts freely on every pair in $Y$: if $\lambda a\equiv a$ and $\lambda b\equiv b$, multiplying the Bezout identity by $\lambda -1$ gives $\lambda \equiv 1(\mathrm{mod}n)$. Hence every equivalence class has size $\phi (n)$. Dividing the answer from part (a) by this size gives

Assume $ad\equiv bc(\mathrm{mod}n)$ and choose $r,s$ as in part (b). Put $\lambda =rc+sd$. Multiplying $ra+sb\equiv 1$ by $c$ and using $ad\equiv bc$ givesSimilarly, multiplication by $d$ gives $\lambda b\equiv d(\mathrm{mod}n)$.

Moreover, $\lambda$ is a unit modulo n: if a prime divided both $\lambda$ and $n$, the two congruences would make it divide $c,d,n$, contrary to $(c,d)\in Y$. Thus $R$ means precisely that two pairs differ by multiplication by a unit modulo $n$. Reflexivity, symmetry using ${\lambda }^{-1}$, and transitivity using products of units now show that $R$ is an equivalence relation.

The condition $\gcd (a,b,n)=1$ says that the ideal of $\mathbb{Z}$ generated by $a,b,n$ is all of $\mathbb{Z}$. The multi-integer Bezout identity therefore supplies integers $r,s,t$ such thatEquivalently, first apply Bézout's identity to $a,b$ and then to $\gcd (a,b),n$.

The inclusion-exclusion principle states that for finite sets $A_1,\dots ,A_k$,For each prime $p\mid n$, exclude pairs whose two coordinates are divisible by $p$. For a set of distinct prime divisors $p_1,\dots ,p_j$, exactly $n^2/(p_1\cdots p_j{)}^2$ pairs have both coordinates divisible by their product. Inclusion-exclusion therefore gives

We prove both identities by mathematical induction. They hold at $n=1$. Assume they hold at $n$. Using $F_{n-1}=F_{n+1}-F_n$,Adding this to $F_{2n+1}=F_n^2+F_{n+1}^2$ and using $F_{n+2}=F_n+F_{n+1}$ givesThese are precisely the two assertions with $n$ replaced by $n+1$, completing the induction.

Fermat's little theorem states that for a prime $p$ and integer $a$,if $p\nmid a$, equivalently $a^{p-1}\equiv 1(\mathrm{mod}p)$. For the latter form, multiplication by $a$ permutes the nonzero residue classes modulo $p$. Multiplying the congruences $a,2a,\dots ,(p-1)a$ and cancelling $(p-1)!$ gives $a^{p-1}\equiv 1$. The first form also covers $p\mid a$.

Since $803=50\cdot 16+3$, the theorem gives

Apply part (a) first to the position $\mathbf{r}$:Applying it again to both terms on the right giveswhere $(\dot{\mathbf{\omega }}{)}_S=(\dot{\mathbf{\omega }}{)}_{S^{\prime }}$ because $\mathbf{\omega }\times \mathbf{\omega }=0$. Substituting the first formula into the last term produces

In the Earth-fixed rotating reference frame, the sideways apparent force is the Coriolis acceleration multiplied by the ski-doo's mass:At the South Pole, $\mathbf{\omega }$ is vertical and antiparallel to the local outward normal, while $\mathbf{v}$ is horizontal, soThe force points to the left of the direction of travel, which is westward when the ski-doo is moving north after crossing the pole.