Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics I / 6

Let $|G|=p$, with $p$ a prime number, and choose $g\ne e$. By Lagrange's theorem, the order of a group element $g$ divides $p$ and is greater than one. It is therefore $p$, and $G=⟨g⟩$. Hence a group of prime cardinality is a cyclic group.

The converse is false: the cyclic group $C_4$ has four elements, and $4$ is not a prime number.

If $G=⟨g⟩$, any two elements are $g^a,g^b$, andThus every cyclic group is an abelian group.

For the converse, the Klein four-group $C_2\times C_2$ is an abelian group, with componentwise multiplication. Each nonidentity element has order of a group element two, so none generates its four elements. It is therefore not a cyclic group.

The hypothesis implies $x^{-1}=x$ for every $x\in G$, including $e$. For $a,b\in G$,Thus $G$ is an abelian group. This is the group of exponent two is abelian argument.

The converse is false: the cyclic group $C_3$ is an abelian group, but its two nonidentity elements have order of a group element three.

Choose distinct $a,b\ne e$, possible because $|G|>2$. By iii, $a$ and $b$ commute, and $a^2=b^2=e$. The four elementsare distinct: $ab=e$ would give $a=b$, while $ab=a$ or $ab=b$ would make $b=e$ or $a=e$. Commutativity and $a^2=b^2=e$ show that products stay in $H$, and every element is its own inverse. More explicitly,is a group isomorphism, with addition of exponents modulo two. Hence $G$ contains a subgroup isomorphic to the Klein four-group.

The converse is false. The direct product of groups $C_2\times C_2\times C_3$ contains $C_2\times C_2\times \{e\}$, but it has elements of order of a group element three. Thus the existence of the specified subgroup does not force every nonidentity element to have order of a group element two.

The complete classification up to group isomorphism isHere $C_1$ is the trivial group. The prime number cases follow from i. For four elements, an element of order of a group element four generates $C_4$. If there is none, Lagrange's theorem makes every nonidentity element have order of a group element two. Then iv produces a four-element subgroup, which must be all of $G$, giving the Klein four-group.

For six elements, an element of order of a group element six gives $C_6$. Suppose there is none. If there were no element of order of a group element three either, all nonidentity elements would have order of a group element two. Part iv would then produce a subgroup of cardinality four, contradicting Lagrange's theorem. Thus choose $a$ of order of a group element three. There is also an involution $b$: otherwise the five nonidentity elements could be paired into distinct pairs $\{g,g^{-1}\}$, impossible for an odd number of elements.

The subgroup $H=⟨a⟩$ has index of a subgroup two, so it is a normal subgroup. Indeed, for $g\notin H$, both $gH$ and $Hg$ are the complement of $H$, and for $g\in H$ both equal $H$. Consequently $ba{b}^{-1}$ is either $a$ or $a^{-1}$. The first possibility makes $a,b$ commute, and $ab$ then has order of a group element six: $(ab{)}^k=e$ forces $a^k=b^{-k}\in ⟨a⟩\cap ⟨b⟩=\{e\}$, so both $3$ and $2$ divide $k$. This contradicts our assumption. ThereforeEvery element is uniquely $a^i$ or $a^{i}b$, with $0\le i<3$, and the relation $ba=a^{-1}b$ determines all their products. Sending $a$ to $(123)$ and $b$ to $(12)$ gives a group isomorphism with the symmetric group $S_3$. This completes the classification of groups of order at most six.

Of these groups, exactly $C_1,C_2,C_3,C_4,C_2\times C_2,S_3$ embed in $S_4$. Concrete subgroups areThere is no $C_5$ subgroup, since $5$ does not divide $24$. There is no $C_6$ subgroup either: the order of a group element in a symmetric group is the least common multiple of the lengths in its disjoint cycle decomposition, and on four letters the only possibilities are $1,2,3,4$.

Label the vertices of a regular tetrahedron $1,2,3,4$, and include all its Euclidean isometries, both rotations and reflections. The following list describes every subgroup of cardinality at most six, allowing arbitrary relabelling of the vertices. The numbers count actual subgroups, rather than just their group isomorphism types.

Here is why this is exhaustive. Part b leaves only the trivial group, $C_2,C_3,C_4,C_2\times C_2,S_3$. The disjoint cycle decomposition gives exactly the two listed kinds of involutions, the three-cycles, and the four-cycles, accounting for all the cyclic subgroups. In a Klein four-group, the nonidentity elements commute. If it contains $(12)$, any other nonidentity element must be a commuting involution, hence $(34)$ or $(12)(34)$; either choice yields the displayed opposite-edge subgroup. If it has no transposition, its three nonidentity elements must be all three double transpositions, yielding the rotational subgroup.

Finally, an $S_3$ subgroup has a unique cyclic subgroup of cardinality three. Its two three-cycles fix the same unique vertex. Every element of the $S_3$ subgroup preserves this vertex, since conjugation preserves that unique cyclic subgroup. Thus it is the full stabilizer subgroup of that vertex. The counts in the list are consequently exhaustive, givingThe two kinds of $C_2$ and the two kinds of Klein four-groups have different geometric behavior despite having the same abstract group isomorphism type within each pair. This is the distinction recorded by the small subgroups of tetrahedral symmetry.