Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2025 / Paper 2 / 8

For the first Cayley table, associativity fails: $(a\ast a)\ast b=1\ast b=b$, whereas $a\ast (a\ast b)=a\ast c=d$. It therefore does not define a group.

For the second table, assign the residues $0,1,2,3,4$ to $1,a,c,b,d$, respectively. Every table entry agrees with addition modulo an integer modulo five. Thus this table defines the cyclic group of order five, with identity $1$ and inverses $a^{-1}=d$, $c^{-1}=b$. The identification with a known group verifies associativity as well as closure and the inverse axioms; inspecting only the identity row and inverses would not suffice.

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.

Write the dihedral group as $D_6=⟨r,s\mid r^6=s^2=1,srs=r^{-1}⟩$, where $r$ rotates through $\pi /3$. There are six rotations and six reflections, so the order is 12. Conjugation by $s$ reverses a rotation, and conjugation by $r$ sends $r^{j}s$ to $r^{j+2}s$. Consequently the conjugacy classes and their sizes areThe last two classes distinguish the reflection axes through opposite vertices and through opposite edge midpoints. Their sizes sum to 12.