Past exam of the algebra course of the University of Oxford / 2010 / B

For the Abelian case of Cauchy theorem for groups, use induction on $|G|$. Choose $a\ne e$. If $p$ divides $m=|a|$, then $a^{m/p}$ has order $p$. Otherwise $⟨a⟩$ is a nontrivial normal subgroup of order prime to $p$, and $G/⟨a⟩$ has smaller order still divisible by $p$. Induction gives a coset of order $p$; any lift $b$ has order divisible by $p$, because the order of its image divides its own order. A suitable power of $b$ then has order $p$. This uses no classification of finite Abelian groups.

For a finite p-group, the class equation iswhere the sum is over noncentral conjugacy classes. Each index is a positive power of $p$, hence divisible by $p$. Therefore $p$ divides $|Z(G)|$, and the center of a group is nontrivial.

If $|G|=p^n$ with $n>1$ and $Z(G)\ne G$, this is already a proper nontrivial normal subgroup. If $Z(G)=G$, the group is Abelian; the established case of Cauchy theorem for groups gives a subgroup of order $p$, which is normal and proper. Thus

If $|G|=p^{a}m$ with $p\nmid m$, a Sylow p-subgroup has order $p^a$. The Sylow theorems say that such subgroups exist; every p-subgroup is contained in one; they are all conjugate; and their number $n_p$ divides $m$ and satisfies $n_p\equiv 1(\mathrm{mod}p)$.

Suppose $|G|=p^{2}q$ with distinct primes and $G$ were simple. A Sylow subgroup of order $q$ has order $q$, and $n_q>1$ divides $p^2$, so $n_q=p$ or $p^2$. If $n_q=p$, conjugation gives a nontrivial homomorphism $G\to S_p$. Its kernel is normal and, by simplicity, trivial. Hence $p^{2}q$ would divide $p!$ by Lagrange's theorem, impossible because $p!$ has only one factor of $p$.

If $n_q=p^2$, distinct Sylow subgroups of order $q$ meet only in the identity and together contain $p^2(q-1)$ nonidentity elements. Only $p^2$ elements of $G$ remain, including the identity. Every Sylow p-subgroup lies entirely in that remaining set and itself has $p^2$ elements. They must therefore all be the same subgroup, making it normal, again a contradiction. Thus no group of order $p^{2}q$ is simple. If $p=q$, the preceding p-group argument applies.

For distinct $p,q$, a finite Abelian group is the direct product of groups of its primary subgroups. A group of order $p^2$ is either cyclic or has every nonidentity element of order $p$, in which case choosing two independent elements gives $C_p\times C_p$. Consequently the possibilities areIf the primes coincide, the Fundamental theorem of finitely generated abelian groups gives instead $C_{p^3}$, $C_{p^2}\times C_p$, and $C_p^3$. This classification is used only for the requested list, not for the proof above.

First establish the two general claims about polycyclic groups. Intersect a subnormal cyclic-factor series $(G_i)$ with $H\le G$. Then $H\cap G_i⊴H\cap G_{i+1}$, and the inclusion induces an injective homomorphismA subgroup of a cyclic group is cyclic, so this is a cyclic-factor subnormal series for $H$, after omitting repetitions. This proves subgroups of polycyclic groups are polycyclic.

To prove the generator bound, choose $g_i\in G_i$ whose coset generates $G_i/G_{i-1}$ whenever that factor is nontrivial. Inductively $G_i=⟨G_{i-1},g_i⟩$, so $G=⟨g_1,\dots ,g_n⟩$. This is the generator bound from a cyclic-factor series: at most $n$ generators suffice.

For the first example, take $G=C_2^m$ with its coordinate flagIt has length exactly $m$ and each factor is $C_2$, so all factors are cyclic.

TakeStart with $1<A_3<S_3$, whose factors are $C_3$ and $C_2$, and then append the $m-2$ coordinate factors of the elementary Abelian group $C_2^{m-2}$. More explicitly, append $S_3\times C_2^j$ for $1\le j\le m-2$. Each subgroup is normal in the next, and the factors consist of one $C_3$ and $m-1$ copies of $C_2$. Thus there are exactly $m$ finite cyclic factors. The group is non-Abelian because it contains $S_3$.

Take the free Abelian groupIts coordinate flag $0<\mathbb{Z}<{\mathbb{Z}}^2<\cdots <{\mathbb{Z}}^m$ is a subnormal series of length $m$, with every factor isomorphic to the infinite cyclic group $\mathbb{Z}$. Hence it has exactly $m$ infinite cyclic factors.

In a finite group, choose a maximal chain of normal subgroups from $1$ to $G$. Finiteness ensures such a chain exists, and maximality leaves no normal subgroup of $G$ strictly between successive terms. This is a chief series.

The infinite cyclic group has no chief series. In any finite chain starting at zero, the first nonzero subgroup is $n\mathbb{Z}$. But $2n\mathbb{Z}$ is a nonzero proper subgroup of it and is normal in $\mathbb{Z}$. Thus zero is not maximal among the normal subgroups in that first term, contradicting the definition.

For any chief series, $G_{n-1}$ is a maximal proper normal subgroup of $G$. The subgroup correspondence theorem identifies normal subgroups of $G/G_{n-1}$ with normal subgroups of $G$ containing $G_{n-1}$. Consequently the top chief factor is simple.

Other chief factors need only have no proper nonzero subgroup invariant under the action of the whole group. In $A_4$,is a chief series, where $V_4=\{1,(12)(34),(13)(24),(14)(23)\}$. Conjugation in $A_4$ permutes the three involutions transitively, so none of its three order-two subgroups is normal in $A_4$. Thus $V_4$ is a minimal nontrivial normal subgroup; the top quotient is $C_3$. But the bottom chief factor $V_4/1\cong C_2\times C_2$ is not simple, since it has proper order-two normal subgroups as an abstract group.

A complex character of a representation of a finite group $G$ is the functionwhere $\rho :G\to \mathrm{GL}(V)$ is a finite-dimensional complex group representation. The character is a class function because trace is unchanged by conjugation. Its degree is $\chi (1)=\dim V$.

The row orthogonality relations for a character table assert that distinct complex irreducible characters are orthogonal and each has squared norm one:Using the linearity and conjugate-linearity of the character inner product,Since the multiplicities are nonnegative integers, this is

By the Maschke theorem, the restricted representation is a direct sum of irreducible $H$-modules. The row orthogonality relations for a character table identify the multiplicity of ${\psi }_i$ as $d_i=⟨\chi {\downarrow }_H,{\psi }_i{⟩}_H$. ThereforeUsing the preceding character inner product identity and the irreducibility of $\chi$,This is the character restriction norm bound. When $[G:H]=2$, the nonzero sum of squares of integer multiplicities is either one or two. Hence the restriction is irreducible, or is the sum of two distinct irreducible characters, each with multiplicity one. This proves the requested index-two character restriction dichotomy directly; normality of an index-two subgroup is a general group theory fact.

Use the column order $1,(123),(12)(34),(12345),(13452)$ in $A_5$. The two restrictions of degree four coincide and equal ${\theta }_4=(4,1,0,-1,-1)$; those of degree five coincide and equal ${\theta }_5=(5,-1,1,0,0)$. The centralizer orders giveThus both are irreducible characters. The remaining given restriction is $\eta =(6,0,-2,1,1)$, andIt splits as two distinct irreducible characters. Neither is $1$, ${\theta }_4$ or ${\theta }_5$, since their character inner products with $\eta$ are zero. A finite group has as many complex irreducible characters as conjugacy classes, so these five exhaust them. The sum of squares of irreducible degrees is $|A_5|=60$. The first three degrees contribute $1+16+25=42$, leaving $18$ for the two constituents, whose degrees sum to six. Hence both have degree three.

Write one constituent as $\theta =(3,x,y,u,v)$. All five conjugacy classes are closed under inversion, so all its values are real: $\chi (g^{-1})=\overline{\chi (g)}$. Orthogonality with $1$, ${\theta }_4$ and ${\theta }_5$ givesSolving gives $x=0$, $y=-1$ and $u+v=1$. The squared norm one gives $9/60+1/4+(u^2+v^2)/5=1$, so $u^2+v^2=3$ and $uv=-1$. Thus $u,v$ are the roots of $t^2-t-1$, namely $\phi =(1+\sqrt{5})/2$ and $\bar{\phi }=(1-\sqrt{5})/2$. Here the bar labels the other real root, rather than complex conjugation. The other constituent is $\eta -\theta$, so the two last values are interchanged.

The Character table of the alternating group A5 is thereforeThe conjugacy class sizes are $1,20,15,12,12$. A normal subgroup is a union of entire conjugacy classes containing the identity, and its order divides $60$ by Lagrange's theorem. A nontrivial such union has size at least $13$; the only proper divisors of $60$ at least $13$ are $15,20,30$. Subtracting the identity would require a sum of some of $20,15,12,12$ equal to $14,19,29$, respectively, none of which is possible. Thus $A_5$ is simple.