Section A

1. Up to isomorphism, there are five groups of order 8:

C8 ,    C2  ⇥ C2  ⇥ C2 ,    C2  ⇥ C4 ,    D8 ,    Q8 .

(a) How many elements of order 4 does each of these groups contain?

They contain 2,0,4,2,6 elements of order 4 respectively.

(b) List all elements of order 2 in the dihedral group of order 8:

D8  = hx,y | x4  = 1 = y , yx = x2 − 1yi.

The elements of order 2 are x2 ,y,xy,x2y,x3y .

(c) Let H = hxyi 6 D8 . Prove that H is not a normal subgroup of D8 .         We have y(xy)y− 1  = yx = x3y 62 {1,xy} = hxyi. So H is not closed under conjugation and hence not normal.

(d) Is the statement (†) below true or false? (Give a proof or a counterexample as appropriate.)

If a,b 2 D8  and ha,bi = D8  then |hai| · |hbi| = 8.      (†)

This is false, D8  = hy,xyi while |y| · |xy| = 2 · 2 = 4.

(e) Which of the groups of order 8 have a normal subgroup H  C4 ?

Since any copy of C4  in a group of order 8 has index 2 it is normal. All of the groups of order 8 contain copies of C4 except C2(3): hx2 i 6 C8 , h(1,y)i 6 C2 ⇥C4 , hxi / D8  and hii / Q8 .

(f) Let G be a group and define · : G ⇥ G ! G to be the left group action of congugation, so g · h = ghg − 1 .

(i) For each group of order 8 determine the set of fixed points of this action. The set of fixed points is the centre of G, thus this is the whole of G i↵ G is abelian. In the non-abelian cases we have Z(D8 ) = {1,x2 } and Z(Q8 ) = {1, −1}.

(ii) Let G be a non-abelian group of order 8 and let H be the set of fixed points of this action. Explain why the quotient group G/H is well-defined and determine which group this is (up to isomorphism).

The centre of any group is a normal subgroup (Prop 9.12 from notes) and hence the quotient group is well-defined. Moreover it has order |G|/2 = 4. We also know (Prop 9.13 from notes) that this quotient is cyclic only if G is abelian, thus the quotient is the unique non-cyclic group of order 4:

C2  ⇥ C2 .

(g) Prove that exactly one of the non-abelian groups of order 8 is isomorphic to a semidirect product C4  o C2 .

We show first that Q8  is not such a semidirect product. To see this note that the normal copies of C4  in Q8  are hii, hji, hki. To be a semidirect product we would need a copy of C2  in Q8  meeting one of these subgroups trivially. But this would need to be generated by an element of order 2, and the only element of order 2 is −1, which lies in all of these subgroups. Hence Q8  is not such a semidirect product.

We have D8    C4  o C2 . To see this take C4    hxi / D8  and C2    hyi 6 D8 and note that these subgroups meet trivially and |D8 | = |hxi| · |hyi| hence we recognise a semidirect product.

2. Consider the following quotient rings:

F3 [x]/(x2 +1),    F5 [x]/(x2 +1),    F5 [x]/(x2 +2),    F5 [x]/(x2 +3).

(a) Prove that exactly three of these rings are fields.

(b) Find the order of these fields and decide which of them are isomorphic.

(c) Describe all field isomorphisms between any pair that are isomorphic.

We know that the quotient ring Fp [x]/q(x) will be a field i↵ q(x) is irreducible over Fp , which in the case of quadratic polynomials is equivalent to having no roots in Fp . Hence we see that x2 +1 is irreducible over F3  but not over F5 , since 22 +1 = 5 = 0 in F5 . Also x2 + 2 and x2 + 3 have no roots in F5  (the only squares in F5  are 0, 1 and 4). Hence they are all fields except F5 [x]/(x2 +1) which fails to be an integral

domain: (x +2)(x +3) = x2 +5x +6 = x2 +1 = 0.

These fields have orders 9, 25 and 25 respectively. Thus only the last two can be isomorphic.

We want to describe all field isomorphisms φ : F5 [x]/(x2 +2) ! F5 [x]/(x2 +3).   Since the elements of these fields are (equivalence classes) of polynomials of the form ax + b with a,b 2 F5  and φ(ax + b) = aφ(x)+ b, φ(0) = 0 and φ(1) = 1 so we need φ to be the identity on F5 . Hence φ is determined by the image of x.

Suppose that φ(x) = cx + d, where c,d 2 F5 . For a bijection we need c  0 (since φ(F5 ) = F5 ). We also require φ(x2 + 2) to be zero, hence we need (cx + d)2 + 2 = e(x2 + 3) for some e 2 F5 . Thus we have c2 x2 + 2cdx + d2 + 2 = ex2 + 3e. Hence d = 0 and 2 = 3e so e = 4. So c2  = 4. Hence c = 2 or c = 3.

Hence we have two field isomorphisms φ1 (ax+b) = 2ax+b and φ2 (ax+b) = 3ax+b.

Section B

3 .    (a)  Classify all groups of order 2021 up to isomorphism .

We have 2021 = 43 · 47 so applying Sylow for p = 43 we have n43  = 1mod43 and n43  | 47 .  Hence n43  = 1 .  Similarly n47  = 1mod47 and n47  | 43 so n47  = 1 . Hence  we  have  unique  subgroups  H,K and  of order  43  and  47  respectively. Uniqueness implies normality  (since conjugate subgroups are isomorphic and hence  in  this  case  equal) .   Moreover,  since  the  only  group  of  order p,  for p prime, is the cyclic group we have H  C43  and K  C47 .

Since  subgroups  of coprime  order  can  only  meet  trivially,  so  H \ K =  {1}. Finally  |G| =  |H| · |K| so  we  now  recognise  a  direct  product  and  so  G   C43  ⇥ C47   C2021  (since any direct product of cyclic groups of coprime orders is cyclic) .

(b)  Prove that if G is a group of order p2 q, where p,q are distinct primes, then G is not simple .

A group is simple i↵ it contains no proper non-trivial normal subgroup, so we need to prove that G contains a proper non-trivial normal subgroup .  If we find a normal subgroup of order p2  or q this will suffice .

We  apply  Sylow  for p and  q .   If either  np   =  1  or  nq   =  1  we  have  a  unique subgroup of order p2  or q which is necessarily normal .  Hence we may suppose that np  > p + 1 and np  | q, as well as nq  > q + 1 and nq  | p2 .  Hence we must have p < q < p2 .

Since  nq   > q + 1  > p and  nq   | p2  we  must  have  nq   = p2 .   Thus we  have p2 distinct copies of Cq  in G.  Such subgroups are either equal or meet trivially, hence G contains p2 (q − 1) elements of order q .  But this implies that there are only p2  elements in G that do not have order q .  Hence there is a unique Sylow p-subgroup of order p2  which is normal in G.

(c)  Classify all groups of order 242 that contain a cyclic subgroup of order 121 .    Let G be a group of order 242 = 2 · 112  containing a subgroup K  C121 .  Since K has index 2 in G it is normal .  Cauchy’s theorem implies that G contains an element of order 2 that generates Q  C2 .

Since  gcd(121 , 2)  =  1  so  K \ Q =  {1}.   Moreover  |G| =  |K| · |Q| hence  we recognise a semidirect product:  C121  o C2 .

Since Aut(C121 )  C110  contains a unique element of order 2 there is a unique non-trivial semidirect product C121  o C2 .  We can describe this explicitly as

hx,y | x121  = 1 = y , yx = x2 − 1yi

which we recognise  as  D242 .   The  other possibility  is  a trival  homomorphism which gives the direct product:  C121  ⇥ C2   C242 .

4 .  Let p be an odd prime and let P 2 Sylp (Sp ) be a Sylow p-subgroup of the symmetric group Sp .

(a)  Prove that P  Cp .

Since  |Sp | = p! =  (p − 1)!p and p - (p − 1)! so a  Sylow p-subgroup of Sp  has order p and hence is isomorphic to Cp  .

(b)  Compute the number of distinct Sylow p-subgroups of Sp .

Any  subgroup  of order p  in  Sp  is  generated  by  a p-cycle  of which  there  are (p− 1)!, moreover each such subgroup contains exactly p − 1 generators .  Hence Sp  contains exactly  (p − 2)! Sylow p-subgroups .

(c)  Let X  = Sylp (Sp ) .  Let  · : Sp  ⇥ X  ! X  be the left group action of Sp  on X  by

conjugation, so σ · H = σHσ− 1 .

(i)  Prove that this is a left group action .

That this is a well-defined map follows from Sylow (2) which tells us that Sylow p-subgroups are conjugate .  To see that it satisfies the two axioms of a left group action we note that for any Sylow p-subgroup H, conjugation by the identity gives  H .   Moreover  compatibility  also holds:  if σ , ⌧  2 Sp

then

σ · ⌧ · H = σ(⌧ H⌧− 1 )σ− 1  = (σ⌧ )H(σ⌧ )− 1  = (σ⌧ ) · H.

(ii)  Prove that Q(P) = {σ 2 Sp  | σ · P = P} is a subgroup of Sp .

This is simply the stabilizer subgroup of P .

(iii)  Compute the order of Q(P) .

By  the  Orbit– Stabilizer  theorem  we  know  that  |Q(P)| =  |Sp |/|Sp  · P|, where Sp · P is the orbit of P under this action .  But Sylow (2) tells us the orbit is all Sylow p-subgroups .  Hence  |Q(p)| = p!/(p − 2)! = p(p − 1) .

(d)  Let n  > 1 be the smallest integer such that Sn  contains a non-abelian Sylow p-subgroup .  Find n in terms of p justifying your answer carefully.                         The  smallest value  of n  for which  Sn  has  a  non-abelian  Sylow p-subgroup  is n = p2 .  To see this note first that if n < p2 , say n = kp+r, for some 0 6 r < p

and  k  > 0  then pk   | n!  but pk+1  - n! .   Hence  a  Sylow p-subgroup  of Sn  has order pk .  If σ 1 , . . . , σk  are  k  disjoint p-cycles in  Sn  then  hσ1 , . . . , σk i  Cp(k)  is an abelian subgroup of Sn  of order pk .  Since Sylow p-subgroups are conjugate and hence isomorphic so all Sylow p-subgroups of Sn  are abelian .

To see that Sp2   has a non-abelian Sylow p-subgroup we note that a Sylow p- subgroup in Sp2   must have order pp+1, since p2 ! = pp+1m, with p - m .  As above we have K =  hσ1 , . . . , σp i  Cp(p)  6 Sp2 .  Sylow’s theorem tells us that this is a proper subgroup of a Sylow p-subgroup H .

To prove that  H  cannot be  abelian  it  is  sufficient to  show that there  are  no permutations in Sp2   − K that commute with all elements of K .

For any fixed p-cycle σ the only permutations that commute with σ are those of the form σj ⌧ where ⌧  is disjoint from σ . Thus if a permutation commutes with all of the p-cycles generating K  then it must belong to K . (Since the generators of K partition the set {1, 2, . . . ,p2 }.)

Hence there are no permutations in Sp2   − K that commute with all elements of K, thus as H contains K as a proper subgroup H must be non-abelian.