Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 3 / 1

The derived series is $G^{(0)}=G$, $G^{(i+1)}=[G^{(i)},G^{(i)}]$, where the bracket denotes the commutator subgroup. The group is soluble ifEquivalently it has a finite series with abelian factors. The trivial group is included.

If $H\le G$, induction gives $H^{(i)}\le G^{(i)}$, because commutators of elements of a subgroup are also commutators in the larger group. Thus termination of the derived series of $G$ forces termination for $H$.

For a normal subgroup $N$, the quotient map sends $[x,y]$ to the commutator of their images. Surjectivity then givesConsequently subgroups and quotient groups of a soluble group are soluble. The assertion about a quotient uses a normal subgroup; it is not a quotient by an arbitrary subgroup.

Let $N\ne 1$ be a minimal normal subgroup. Its commutator subgroup $N^{\prime }$ is characteristic in $N$, hence normal in $G$. Minimality makes $N^{\prime }=1$ or $N^{\prime }=N$. The latter would prevent the soluble group $N$ from having a terminating derived series, so $N^{\prime }=1$ and $N$ is abelian.

Choose a prime $p$ dividing $|N|$. In a finite abelian group its Sylow $p$-subgroup is characteristic, so minimal normality makes this subgroup all of $N$. The subgroup ${\Omega }_1(N)=\{n\in N:n^p=1\}$ is nontrivial, characteristic and hence normal in $G$. Minimality again makes it all of $N$. Thusan elementary abelian p-group. Both abelianness and minimal normality are essential to the two characteristic subgroup arguments.

A Hall subgroup for a prime set $\pi$ is a subgroup $H\le G$ whose order has only prime divisors in $\pi$ and whose index has no prime divisor in $\pi$:Here one is allowed in either class, and ${\pi }^{\prime }$ is the complementary set of primes. Equivalently $|H|$ contains the complete prime-power contribution to $|G|$ for each prime in $\pi$.

We prove Hall subgroup existence in soluble groups by induction on $|G|$. The trivial group is immediate. Choose a nontrivial minimal normal subgroup $N$, elementary abelian of order $p^a$ by part (c). Induction gives a Hall $\pi$-subgroup of $G/N$; let $K$ be its full preimage.

If $p\in \pi$, then $K$ itself is the required subgroup. If $p\notin \pi$ and $K<G$, apply induction inside the soluble subgroup $K$ to obtain a Hall $\pi$-subgroup $H$ of $K$. Since $[G:K]$ and $[K:H]$ are both ${\pi }^{\prime }$-numbers, $H$ is Hall in $G$ too.

It remains to treat $p\notin \pi$ and $K=G$. Then $G/N$ is a $\pi$-group. If $G=N$, the subgroup one works. Otherwise choose a minimal normal subgroup $M/N$ of $G/N$, an elementary abelian $q$-group with $q\in \pi$. Let $Q$ be a Sylow $q$-subgroup of $M$. As $|M|=|N|q^b$ and $p\ne q$, we have $M=NQ$. The permitted Frattini argument givesThe last equality uses $Q\le N_G(Q)$ and the normality of $N$.

If $T=N_G(Q)<G$, then $[G:T]=[N:N\cap T]$ is a power of $p$, hence a ${\pi }^{\prime }$-number. Apply induction to $T$ and multiply indices as before. If $T=G$, then $Q$ is a nontrivial normal $\pi$-subgroup. Induction in $G/Q$ gives a Hall $\pi$-subgroup whose full preimage in $G$ is Hall, since its additional factor $|Q|$ is a $\pi$-number. These cases exhaust the possibilities, proving existence for every prime set. No unproved complement theorem or conjugacy theorem for Hall subgroups was inserted into the proof.

Use the Sylow theorems. The number $n_p$ divides $qr$ and is congruent to one modulo $p$. Since $q,r<p$, it is either one or $qr$. Suppose $n_p=qr$. Different subgroups of prime order intersect trivially, so their nonidentity elements occupy $qr(p-1)$ places, leaving only $qr-1$ nonidentity elements of other prime orders.

If neither the Sylow $q$-subgroup nor the Sylow $r$-subgroup is normal, then $n_q\ge p$ and $n_r\ge q$. Indeed $n_q$ divides $pr$, and its possible divisor $r<q$ cannot satisfy the Sylow congruence; the smallest remaining nontrivial possibility is at least $p$. Likewise any nontrivial divisor of $pq$ is at least $q$. Their elements would require at leastplaces, since the excess is $(p-1)(q-1)>0$. Hence some Sylow subgroup of order $s\in \{q,r\}$ is normal.

In the quotient by this normal subgroup, the largest prime $p$ has a normal Sylow subgroup: for a group of order $pt$ with $p>t$ its Sylow count divides $t<p$ and so equals one. Pulling back gives a normal subgroup of order $ps$. Inside it, the subgroup of order $p$ is again the unique Sylow $p$-subgroup. It is characteristic in that normal subgroup and therefore normal in $G$, contradicting $n_p=qr$. Thus $\boxed{n_p=1}$.

Let $P$ be this normal Sylow subgroup. In $G/P$, of order $qr$, its subgroup of order $q$ is normal by the same argument. Its preimage $K$ is a normal Hall $\{p,q\}$-subgroup. The serieshas factors of orders $p,q,r$, hence cyclic and abelian. Therefore $\boxed{G\text{ is soluble}.}$ This establishes solubility before using any Hall-existence conclusion that itself assumes solubility.