Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 2 / 8

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 operation is not defined for every pair in the proposed set: choosing $a=1,b=-1$ makes its denominator zero. Thus it is not even a binary operation on $\mathbb{C}\setminus \{0\}$, and is not a group. In particular, removing zero from the domain does not prevent the sum in the denominator from vanishing.

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.

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 subgroup $H\le G$ and $g\in G$, the left and right cosets are $gH=\{gh:h\in H\}$ and $Hg=\{hg:h\in H\}$. A normal subgroup satisfies $gH{g}^{-1}=H$ for every $g$. Multiplying this set equality on the right by $g$ givesConversely this equality implies $gH{g}^{-1}=H$, so it is an equivalent definition of normality.

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

Write a matrix in the set as $M(a,b)$. The zero matrix $M(0,0)$ has no multiplicative inverse matrix, so $S$ is not a group. Its determinant iswhich vanishes only at that zero matrix. It is therefore the only element that needs to be removed.

For closure, direct matrix multiplication givesThe product of nonzero determinants is nonzero, so two nonzero elements never multiply to zero. The identity matrix is $M(1,0)$, associativity is inherited from matrix multiplication, andThus $\boxed{S\setminus \{0_{2\times 2}\}\text{ is a group}}$. It is the multiplicative group of nonzero quaternions in the complex matrix representation of quaternions.