Past exam of the algebra course of the University of Oxford / 2007 / B

For $|G|=p^{r}m$ with $p\nmid m$, Sylow's theorem gives subgroups of order $p^r$, says that every $p$-subgroup lies in a conjugate of one of them, and that all such Sylow subgroups are conjugate. Their number $n_p$ divides $m$ and satisfies $n_p\equiv 1(\mathrm{mod}p)$. A group is soluble when its derived series, obtained by repeatedly taking the commutator subgroup, reaches the identity. Equivalently it has a subnormal series with abelian factors.

For a group of order $p^2$, the class equation shows its center is nontrivial: noncentral conjugacy classes have sizes divisible by $p$. If the center had order $p$, the quotient by the center would be cyclic; a group with cyclic central quotient is abelian, since representatives are powers of one element times central elements. This would make the center the whole group. Thus every group of order $p^2$ is abelian.

Now consider $|G|=p{q}^a$ with $0\le a\le 2$. Cases with a normal Sylow subgroup are soluble: that subgroup has prime or prime-square order and is abelian, and its quotient also has such order and is abelian. This gives a subnormal series with abelian factors.

If $p<q$, the Sylow number $n_q$ divides $p$ and is $1$ modulo $q$, so is one. If $p>q$ and $a\le 1$, $n_p$ divides $q^a<p$, so again is one. For $p>q$ and $a=2$, a nontrivial $n_p$ can only be $q^2$, since $q<p$. The congruence would force $p\mid (q-1)(q+1)$, hence $p\mid q+1$. Since $p>q$, this means $p=q+1$, and primality forces $(p,q)=(3,2)$.

In the remaining order-twelve case, either a Sylow subgroup is already normal, or there are four Sylow subgroups of order three. Those have pairwise trivial intersection and contribute eight distinct elements of order three. Only four elements remain, including the identity. Every Sylow subgroup of order four lies entirely among those four elements, so it is unique and normal. It and its quotient are abelian. Therefore all groups of the required orders are soluble, by elementary Sylow arguments alone.

A composition series is a chain $1=G_0◃G_1◃\cdots ◃G_r=G$ with simple nontrivial factors. Every finite group has one, by recursively choosing a maximal proper normal subgroup. The Jordan–Hölder theorem states that any two such series have the same length and the same multiset of simple factors up to isomorphism.

For a proof induct on $|G|$. Let $M,N$ be the penultimate terms in two series. If $M=N$, apply induction there. Otherwise their maximal normality gives $MN=G$. Set $D=M\cap N$. The second isomorphism theorem givesso these quotients are simple. Choose a composition series of $D$ and extend it once to $M$ and once to $N$. Induction in the smaller groups $M,N$ compares these with the respective original prefixes. Appending $G/M$ or $G/N$ gives in both cases the factors of $D$ together with $G/M,G/N$. This completes the proof, including the case $D=1$.

Exactly one: take $S_5$. Its only nontrivial proper normal subgroup is $A_5$. Indeed a normal $N$ has $N\cap A_5$ either $1$ or $A_5$ by the supplied simplicity. In the first case $N$ injects into $S_5/A_5\cong C_2$, so a nontrivial $N$ would be a normal subgroup of order two and hence central. But $S_5$ has trivial center: conjugating every transposition trivially forces preservation of each unordered pair, and therefore of every point. Thus the unique series is $1◃A_5◃S_5$.

Exactly two: take $A_5\times C_2$. A normal subgroup containing $A_5\times 1$ is that subgroup or the whole product. For a normal subgroup $N$ meeting $A_5\times 1$ trivially, commutators of its elements with $A_5\times 1$ vanish; its first coordinates therefore lie in $Z(A_5)=1$. Hence $N\subseteq 1\times C_2$. The two nontrivial proper normal subgroups are exactly the two factors, each simple, so the two series pass respectively through $A_5\times 1$ and $1\times C_2$.

More than two: take $C_{120}$. A cyclic group has exactly one subgroup of each divisor order, and composition steps have prime indices. The orders of those indices are the distinct permutations of $2,2,2,3,5$, givingAll three example groups have order $120$.

An extension of $A$ by $B$ is a short exact sequence $1\to A\to G\to B\to 1$. It splits if the quotient map has a homomorphic section. Equivalently the embedded $A$ has a subgroup complement isomorphic to $B$, and $G$ is the corresponding semidirect product.

For $t\notin K$ and $a\in K$, both $t$ and $ta$ lie outside $K$, so $t^2=(ta{)}^2=1$. It follows that $tat=a^{-1}$, and $t=t^{-1}$ gives the claimed conjugation identity. Conjugation is an automorphism of $K$, so inversion is an automorphism there. Comparing the image of $ab$ with the product of the images gives $b^{-1}{a}^{-1}=a^{-1}{b}^{-1}$. Therefore $K$ is abelian.

Conversely inversion is an automorphism of any abelian group $A$, of order dividing two. Form the semidirect product $G=A⋊C_2$ with multiplicationFor completeness associativity follows because $b\mapsto b^{(-1{)}^{\epsilon }}$ is an automorphism and these automorphisms compose according to addition modulo two; $(1,0)$ is the identity and inverses follow from this formula. The subgroup $H=A\times \{0\}$ is normal with quotient $C_2$, and every outside element satisfies $(a,1{)}^2=(a{a}^{-1},0)=1$. Thus the generalized dihedral group supplies the required extension, including when $A$ is infinite or has elements of order two.

The degree-sum formula for irreducible characters givesThe number of linear characters is the order of the abelianization, so $|G/G^{\prime }|=3$ and $|G^{\prime }|=7$. The intersection of the kernels of all linear characters is $G^{\prime }$. Consequently $g_4,g_5$ are nonidentity elements of this order-seven subgroup and have order seven. The images of $g_2,g_3$ under a displayed linear character have order three, so their orders are divisible by three. Lagrange's theorem leaves three or twenty-one; an element of order twenty-one would make $G$ cyclic, contradicting its degree-three irreducible characters. ThereforeThe center is trivial. A nontrivial proper center would have order three or seven, giving a cyclic quotient and hence an abelian group, again a contradiction. Thus a centralizer of a nonidentity element is proper. It has order three for an order-three element, and seven for an order-seven element. The class sizes are therefore $1,7,7,3,3$.

Column orthogonality says ${\sum }_i|{\chi }_i(g){|}^2=|C_G(g)|$. For $g_2,g_3$, the three linear characters already contribute three, so the remaining two entries must both vanish. Orthogonality with the identity column says ${\sum }_i{\chi }_i(1)\overline{{\chi }_i(g)}=0$ for $g\ne 1$. At $g_4$ this gives $1+\alpha +{\chi }_5(g_4)=0$, so ${\chi }_5(g_4)=\beta$, using $\alpha +\beta =-1$. Similarly ${\chi }_5(g_5)=\alpha$. Thus the completed table isHere $\beta =\overline{\alpha }$, $|\alpha {|}^2=|\beta {|}^2=2$ and $\alpha \beta =2$. These relations, together with $1+\omega +{\omega }^2=0$, verify all weighted row orthogonality relations and the centralizer column norms.