Past exam of the mathematics course of the University of Cambridge / 2002 / ia / Paper 1 / Section I

For a finite group $G$ acting on a set $X$, write $G_x=\{g\in G:gx=x\}$ for the stabilizer subgroup and $Gx=\{gx:g\in G\}$ for the orbit of $x$. The orbit-stabilizer theorem givesIndeed, $g{G}_x\mapsto gx$ is a bijection from the left cosets of $G_x$ onto $Gx$: two images agree precisely when $h^{-1}g\in G_x$. Each coset has $|G_x|$ elements, proving the formula.

Place the cube vertices at $(\pm 1,\pm 1,\pm 1)$. The four vertices whose coordinate product is $+1$ form $T$; those with product $-1$ form $T^{\prime }$. Within either set, two distinct vertices differ in exactly two coordinates, so every edge has length $2\sqrt{2}$ and each set is a regular tetrahedron.

A member of the cube rotation group is a signed permutation matrix of determinant one. There are $3!$ permutations of the coordinate axes and four choices of signs for each permutation that give determinant one, so $|G|=24$. Such a matrix maps a coordinate-product class to a coordinate-product class. Moreover, the rotation through a quarter-turn $(x,y,z)\mapsto (-y,x,z)$ interchanges the classes. Thus the orbit of $T$ consists of exactly $T,T^{\prime }$. By the orbit-stabilizer theorem, the tetrahedron stabilizer in the cube rotation group has order

The fundamental theorem of algebra says that every nonconstant polynomial with complex number coefficients has a root in the complex numbers. Repeated polynomial division by a factor of degree one therefore factors a degree-three polynomial with complex number coefficients into three factors of degree one, with algebraic multiplicities included.

For a matrix with complex number entries $A$ of size three, its characteristic polynomial is ${\chi }_A(t)=\det (tI-A)$, and its characteristic equation is ${\chi }_A(t)=0$. A complex number $\lambda$ is an eigenvalue exactly when $\lambda I-A$ is singular, equivalently when its null space contains a nonzero vector. Thus the three roots of the characteristic polynomial are the three eigenvalues counted with algebraic multiplicity.

For the specified matrix, expansion along the first row givesHence the eigenvalues are $1,1,-1$. Corresponding eigenvectors areThe matrix with these three columns has determinant $2i\ne 0$, so these eigenvectors are linearly independent. A repeated eigenvalue does not in general guarantee two independent eigenvectors; here they have been exhibited directly.

The limit of a sequence of real numbers is $a$ if, for every $\epsilon >0$, there is an integer $N$ such that $n\ge N$ implies $|a_n-a|<\epsilon$. It tends to positive infinity if, for every real $M$, there is an integer $N$ such that $n\ge N$ implies $a_n>M$.

If $a_n\to +\infty$, choose the latter threshold with $M=1/\epsilon$. Then $a_n>1/\epsilon >0$ eventually, so $|1/a_n|<\epsilon$. Thus $1/a_n\to 0$. The converse is false: $a_n=-n$ has reciprocal tending to zero and tends to negative infinity. The precise sign-free statement is given by reciprocal limits and escape in absolute value: $1/a_n\to 0$ is equivalent to $|a_n|\to \infty$.

If $a_n\to a\ne 0$, eventually $|a_n-a|<|a|/2$, and the triangle inequality gives $|a_n|>|a|/2$. ConsequentlyGiven $\epsilon >0$, take $N$ large enough that $|a_n-a|<\min (|a|/2,\epsilon |a{|}^2/2)$ for all $n\ge N$. This proves $1/a_n\to 1/a$ directly from the definition of the limit of a sequence.

Here is a nested-interval proof of the Bolzano-Weierstrass theorem. Put every term of the bounded real sequence in a closed interval $I_0=[-M,M]$, with $M>0$. Bisect this interval and choose a closed half containing infinitely many terms. Repeating gives nested closed intervals $I_j=[l_j,r_j]$, each containing infinitely many sequence terms, with length $2M/2^j$.

By the least-upper-bound property, $x_{\ast }={\sup }_j{l}_j$ exists. For each $j$, every lower endpoint lies at most $r_j$, while $l_j\le x_{\ast }$; hence $x_{\ast }\in I_j$. Choose increasing indices $n_j$ with $a_{n_j}\in I_j$, possible because each interval contains infinitely many terms. ThenThus every bounded real sequence has a convergent subsequence.

For a sequence with no convergent subsequence, take $a_n=n$: any subsequence has $n_j\ge j$, so its terms tend to positive infinity rather than to a finite limit of a sequence. For an unbounded sequence with a convergent subsequence, take $a_{2n}=0$ and $a_{2n-1}=n$. Its odd terms are unbounded, while its even subsequence converges to zero.