Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / 9

For $g\in G$, the left coset of $H$ represented by $g$ isIt is a subset of $G$ obtained by multiplying every element of $H$ on the left by the same group element. The identity representative gives $eH=H$.

For each $g$, the mapis surjective by definition of a coset and injective by left cancellation. Its inverse is multiplication by $g^{-1}$. ThereforeThe same argument with right multiplication proves the analogous result for right cosets. This equality of sizes is the counting step in Lagrange's theorem.

Write $G=\{e,a,b,c\}$. By Lagrange's theorem, the order of a group element divides four. Since no element has order four, all three nonidentity elements have order two. Thus every element is its own inverse. For any $x,y\in G$,so the group is Abelian. Also $ab$ cannot equal $e,a$ or $b$, by cancellation and the fact that $a,b$ are distinct nonidentity elements. Hence $ab=c$. The remaining products are forced, givingThe map $e\mapsto (0,0)$, $a\mapsto (1,0)$, $b\mapsto (0,1)$, $c\mapsto (1,1)$ preserves this table under componentwise addition modulo two. Thereforethe Klein four-group, also called the Vierergruppe.

A cyclic group is generated by one element: $G=⟨g⟩=\{g^n:n\in \mathbb{Z}\}$. The hypothesis needs the qualification no proper nontrivial subgroup: every nontrivial group has the proper trivial subgroup $\{e\}$, so the literal wording admits no examples.

Under that intended hypothesis, choose $g\ne e$. The subgroup $⟨g⟩$ is nontrivial and must therefore be all of $G$, proving cyclicity. If $g$ had infinite order, then $⟨g^2⟩$ would be nontrivial and proper, since $g$ is not an even power of an infinite-order generator. Thus $g$ has finite order $N>1$ and $|G|=N$.

If $N$ were composite, choose a divisor $d$ with $1<d<N$. Then $⟨g^d⟩$ would have order $N/d$, strictly between one and $N$, again contradicting the hypothesis. ConsequentlyConversely, Lagrange's theorem shows that a group of prime order has only itself and its trivial subgroup. This proves the characterization of groups with no proper nontrivial subgroups, even without assuming finiteness at the outset.