Past exam of the algebra course of the University of Oxford / 2008 / B

Write $C=⟨c⟩$ with $|c|=n$. For each divisor $k$ the subgroup $⟨c^{n/k}⟩$ has order $k$. Conversely, if $H$ has order $k$, Lagrange gives $k\mid n$, and every $c^j\in H$ satisfies $(c^j{)}^k=1$, so $n\mid jk$. Therefore $n/k\mid j$ and $H\subseteq ⟨c^{n/k}⟩$. Equality follows from the common order. Thus there is exactly one subgroup for each divisor of $n$.

For the Abelian assertion, take a surjection ${\mathbb{Z}}^n\to A$ and let $L$ be the inverse image of $B$. Every subgroup of ${\mathbb{Z}}^n$ has at most $n$ generators, by induction: its projection onto the first coordinate is either zero, reducing to ${\mathbb{Z}}^{n-1}$, or $d\mathbb{Z}$ for some $d>0$. In the latter case choose one element with first coordinate $d$; subtracting its multiples reduces every element to the kernel, generated by at most $n-1$ elements by induction. The base case is the elementary description of subgroups of $\mathbb{Z}$ by their least positive element. Images of these generators generate $B$, proving the claim.

In $S_6$, the commuting disjoint transpositions $(12),(34),(56)$ generate $C_2^3$. This is a three-dimensional vector space over ${\mathbb{F}}_2$, so it requires three generators and cannot be generated by two.

For each $n\ge 3$, take $G=S_{2n+2}$, generated by $(12)$ and the cycle $(12\dots 2n+2)$. Its subgroup generated byis $C_2^{n+1}$. It requires $n+1$ generators by the same dimension argument, giving a two-generated group with a subgroup requiring more than $n$ generators.

A Sylow subgroup for the prime $p$ is a maximal $p$-subgroup; in a finite group of order $p^{a}m$ with $p\nmid m$ it has order $p^a$. Sylow theorems assert existence, containment of every $p$-subgroup in a Sylow $p$-subgroup, conjugacy of all Sylow $p$-subgroups, and the counting conditions $n_p\mid m$, $n_p\equiv 1(\mathrm{mod}p)$.

Since $N$ is a normal subgroup, $NP$ is a subgroup andThe final index is prime to $p$. Since $N\cap P$ is a $p$-subgroup, its index in $N$ being prime to $p$ means it contains the entire $p$-part of $|N|$. Hence $N\cap P$ is Sylow in $N$.

The quotient $PN/N$ is a $p$-group, being the image of $P$, andIts index is prime to $p$, so $PN/N$ is Sylow in $G/N$.

If $P\le A$ and $Q\le B$ are Sylow $p$-subgroups, their product has the full $p$-part of $|A\times B|$, hence is Sylow. Conversely, for any Sylow $R\le A\times B$, its two projections are $p$-subgroups and lie in some Sylows $P,Q$. Then $R\subseteq P\times Q$, and equality follows by order. Distinct pairs give distinct subgroups because their projections recover $P$ and $Q$. Thus

Define $G^{(0)}=G$ and $G^{(i+1)}=[G^{(i)},G^{(i)}]$. A group is a soluble group if this derived series reaches the identity after finitely many steps. Its derived length is the least $d$ with $G^{(d)}=1$.

Let $V=\{1,(12)(34),(13)(24),(14)(23)\}\le A_4$. This Klein four-group is a normal subgroup and Abelian; $A_4/V$ has order three, so $A_4^{\prime }\subseteq V$. The group $A_4$ is nonabelian, so its derived subgroup is nontrivial. Conjugation by a three-cycle permutes the three nonidentity elements of $V$ transitively. Since the derived subgroup is a normal subgroup, containing any one forces it to contain all three. Thus $A_4^{\prime }=V$, and $V^{\prime }=1$. Therefore $A_4$ has derived length two.

If $H\le G$, every commutator in $H$ is one in $G$, and induction gives $H^{(i)}\subseteq G^{(i)}$ for all $i$. Thus a subgroup of a solvable group is solvable, with no larger derived length.

The sign map gives $S_4^{\prime }\subseteq A_4$. The commutator $[(12),(23)]$ is a three-cycle. Normality of $S_4^{\prime }$ supplies every conjugate three-cycle, and three-cycles generate $A_4$. Hence $S_4^{\prime }=A_4$, so the complete derived series isEvery inclusion is strict, proving solvability and derived length three.

For the preliminary index assertion, the coset action gives a homomorphism $G\to S_n$. Its kernel is a normal subgroup and cannot be all of $G$, because the action is transitive on $n>1$ points. Simplicity makes the kernel trivial. The sign homomorphism of this faithful action must be trivial: otherwise simplicity would identify $G$ with $C_2$, impossible for a transitive action of degree at least three. Thus $G$ embeds in $A_n$, and Lagrange givesA group of order $112=16\cdot 7$ has a Sylow two-subgroup of index seven. If the group were simple, the preceding result would give $112\mid 7!/2=2520$. This fails because $112$ is divisible by $16$ while $2520$ has only three factors of two. Thus there is no simple group of order $112$.

For order $56$, Sylow gives $n_7\in \{1,8\}$. If it is one, the order-seven Sylow is a normal subgroup. Otherwise its eight Sylows of order seven intersect only at the identity, giving $8\cdot 6=48$ nonidentity elements of order seven. Only eight elements remain including the identity. Every Sylow two-subgroup has order eight and consists entirely of these remaining elements. Hence all Sylow two-subgroups coincide and the order-eight subgroup is a normal subgroup. Either case gives a nontrivial proper normal subgroup, so no group of order $56$ is simple.

For order $105=3\cdot 5\cdot 7$, Sylow's congruence and divisibility conditions giveIf neither Sylow subgroup were a normal subgroup, there would be $21\cdot 4=84$ elements of order five and $15\cdot 6=90$ elements of order seven. Distinct subgroups of either prime order intersect only at the identity, and elements of the two orders are disjoint. This would give at least $84+90+1=175$ elements in a group of order $105$, a contradiction. Thus a subgroup of order five or seven is a normal subgroup, and the group is not simple.