1.  (a) Give an example of a non-abelian group G with a proper non-trivial normal subgroup H .

S3  is non-abelian and h(123)i  C3  is a proper non-trivial normal subgroup. To see that it is normal we use (b) below. [4 marks]

(b) Prove that if H is a subgroup of a group G and |G/H| = 2 then H is normal in G.

If H has index 2, then since distinct left cosets are disjoint the left cosets are H and G − H . But the same is true for right cosets and hence the unique non-trivial left coset is equal to the non-trivial right coset. Thus H / G. [4 marks]

(c) Give an example of a group G and a subgroup H such that |G/H| = 3 and H is not normal in G.

Let G = S3  and take H = h(12)i. So |H| = 2 and |G/H| = 6/2 = 3. Then (13)H = {(13), (123)}  {(13), (132)} = H(13).

[5 marks]

(d) Let G be a non-abelian group of order 125. Find Z(G) and the quotient group G/Z(G).

If G is non-abelian then Z(G)  G. So since |Z(G)| divides |G| = 125 (by Lagrange’s Thm) so |Z(G)| = 5 or 25. From lectures we saw that G/Z(G)cyclic implies that G is abelian and hence G/Z(G) cannot be cyclic. If |Z(G)| = 25 then |G/Z(G)| = 5 and G/Z(G)  C5 , hence |Z(G)| = 5 and so Z(G)  C5 and G/Z(G) is a non-cyclic group of order 25 of which there is only one: C5  ⇥ C5 . [5 marks]

(e) Prove that if G is a group of odd order and H is a subgroup with |G/H| = 3 then H is normal in G.

Let H be a subgroup of index p in a group G whose order has smallest prime factor p.

Define an action of G on the left cosets of H by x · (gH) = xgH . Labelling the left cosets as L1 ,L2 , . . . ,Lp  this induces a homomorphism from G to Sp  whose kernel is the set K = {g 2 G | gLi  = Li ,  i = 1, . . . ,p}. Since gH = H i↵ g 2 H we must have K ✓ H . Now G/K is isomorphic to a subgroup of Sp so has order dividing |Sp | = p!. It also has order dividing |G|. Since K ⇢ H so |G/K| >  |G/H| = p. Now G is has no prime factors smaller than p so |G/K| = p and hence H = K is normal. [7 marks]

2 .    (a)  Let p > 2 be prime .  Describe all groups of order p.

They are all isomorphic to Cp , since any element of a group of order p generates a  subgroup that  has  order  dividing p,  i .e .   is  either the  identity  or  generates the whole group .  [2 marks]

(b)  Give  two  examples  of  non-isomorphic  groups  of  order p2 ,  explaining  clearly why they are non-isomorphic .

Cp2  and Cp ⇥Cp are non-isomorphic groups of order p2 .  They are not isomorphic since Cp has an element of order p2 , while Cp ⇥ Cp does not .  [2 marks]

(c)  For which pairs of primes p > q is there a unique group of order pq?                   The short answer is there is a unique group of order pq i↵ q - p − 1 .  [4 marks] To see why let G be a group of order pq, where p > q and p,q are both prime . Sylow’s  theorem  tells  us  that  np ⌘ 1modp and  np | q,  hence p > q implies that np = 1 .  So there is a unique subgroup of order p, isomorphic to Cp , which is therefore  normal  in  G.   Sylow  also  implies there  is  a  subgroup  of order  q, isomorphic to Cq .  Since these subgroups meet only in the identity (since their intersection has order dividing both p and  q)  and  |Cp ||Cq | = pq =  |G| so we can recognise a semi-direct product:  G = Cp o✓ Cq , where ✓ : Cq ! Aut(Cp) is a homomorphism .

Now Aut(Cp)  Cp−1  so if q - p − 1 there are no non-trival homomorphisms, and the semi-direct product is a direct product and so G  Cp ⇥ Cq  Cpq .  [4 marks]

If q | p − 1 then there is also a non-abelian group of order pq.  Let Cp =  hx | xp = 1i.  Let φh be a generator for Aut(Cp) = hφhi,  (this exists since Aut(Cp) is  cyclic)  where  2  6 h 6 p − 1  and  φh :  Cp ! Cp is  given  by  φh (x)  =  xh . Then writing Cq = hy | yq = 1i we have a non-trivial homomorphism ✓ : Cq ! Aut(Cp) given by ✓(y) = φh , now

Cp o✓ Cq = hx,y | xp = 1 = yq , yx = xhyi,

is a non-abelian group of order pq.  [4 marks]

(d)  Classify all groups of order 4907 explaining clearly all steps of your argument . We have 4907 = 7 ⇥ 701 and 701 is prime and 7  | 700 .  So using the previous part we know this is a semi-direct product and either it is trivial and we get C4907  or it is non-trivial .

We need to check that the non-trivial semi-direct products are all isomorphic . We have Aut(C701 ) = C700  and we can check that φ2  : C700  ! C700 , φ2 (x) = x2 generates Aut(C700 ) .  (Since 700 = 22 ⇥ 52 ⇥ 7 it is sufficient to verify that none of 2700/2 , 2700/5 , 2700/7  equal 1 modulo 701 .)

We then need to identify all non-trivial homomorphisms ✓ : C7  ! Aut(C701 ) =    C700 . We must have ✓(y) is an element of Aut(C701 )of orderdividing7, soeither    ✓ is trivial or ✓(y)has order7 .  In which case ✓(y) = φ2(100) ,  φ φ φ φ Since φ2(100)  = φ19  this yields six non-abelian groups of the form

hx,y | x701  = 1 = y7 , yx = x19jyi,  j = 1, . . . , 6.  [6 marks]

However y 7! y19 is an automorphism of C7  (it is y 7! y5 ) hence by a result from lectures (Prop 8.5) on isomorphic semi-direct products these are all isomorphic. [3 marks]

3.  (a) Let G be a group of order 24. List all possible orders of the subgroups of G. A group of order 24 could have subgroups of order 1,2,3,4,6,8,12,24. [4 marks]

(b) Using Sylow’s theorem decide what are the possible values of n2  and n3  for G? Since 24 = 23  ⇥ 3 so n2  is the number of distinct subgroups of order 8, while n3  is the number of distinct subgroups of order 3. By Sylow n2  ⌘ 1mod2,

n3  ⌘ 1mod3, n2  | 3 and n3  | 8. So n2  = 1, 3 and n3  = 1, 4. [6 marks] (c) Classify all groups of order 24 with n2  = 1 and n3  = 1.

In this case the subgroups of order 3 and 8 are unique and hence normal. Moreover gcd(3, 8) = 1 so they meet trivially and 24 = 3 ⇥ 8 so G is a direct product of group of order 3 which must be C3  and a group of order 8 which can we any of C8 ,C2  ⇥ C4 ,C2(3),Q8 ,D8  so there are 5 possibilities. [5 marks]

(d) Using results from the course try to classify groups of order 24 with n2  > 1. [Write no more than three pages.]

[This question was beyond the level of examples seen in the course and meant to give those students who had mastered the course material the chance to shine.]

If n3  = 1 then G must contain a unique and hence normal subgroup isomorphic to C3 . Since it also contains a subgroup H of order 8 which meets C3  trivially so we recognise a semi-direct product: G = C3  o✓ H, where H 2 {C8 ,C2  ⇥ C4 ,C2(3),Q8 ,D8 } and ✓ : H ! Aut(C3 ). Note that Aut(C3 )  C2  is generated by φ2  : x 7! x2 . Since n2  = 3 so ✓ must be non-trivial. We consider the five choices of H in turn. Note that since the 2-Sylow subgroups are all conjugate they are isomorphic so distinct choices for H give distinct groups G. The only isomorphisms will be between semi-direct products with the same 2-Sylow subgroups.

H = C8  = hy | y8  = 1i: ✓(y) = φ2  is the only non-trivial homomorphism and yields the group C3  o✓ C8 .

H = C2  ⇥ C4   = hy,z | y2   = z4   = 1, yz = zyi: there are three non-trivial homomorphisms. Writing each in terms of the image of (y,z) we have ✓ 1  = (φ2 ,id), ✓2  = (id,φ2 ), ✓3  = (φ2 , φ2 ). In this case C3  o✓ 1   (C2  ⇥ C4 )  C3  o✓3 (C2  ⇥ C4 ). To see this let f : C2  ⇥ C4  ! C2  ⇥ C4  be defined by f(y) = y, f(z) = yz. Then we can check this defines an automorphism of C2  ⇥ C4  and ✓ 1  = ✓3 ◦ f. So Prop 8.5 from lectures tells us they are isomorphic.

We need to justify why C3  o✓ 1   (C2  ⇥ C4 )  C3  o✓2   (C2  ⇥ C4 ). To see this note that ker(✓1 ) = hzi  C4  while ker(✓2 ) = hy,z2 i  C2  ⇥ C2 . So the former semi-direct product contains elements of order 12, such as xz, while that the later does not.

H = C2(3)  = ha,b,c | a2  = b2  = c2  = 1, ab = ba,ac = ca,bc = cbi: there are seven non-trivial homomorphisms given by deciding whether or not each of a,b,c is mapped to φ2  (at least one of them must be for the homomorphism to be non-trivial). Writing each homomorphism in terms of the images of a,b,c we have ✓ 1  = (φ2 ,id,id), ✓2  = (id,φ2 ,id), . . . , ✓6  = (id,φ2 , φ2 ), ✓7  = (φ2 , φ2 , φ2 ). It is easy to see that C3  o✓ 1   C2(3)    C3  o✓2   C2(3)    C3  o✓3   C2(3)  and C3  o✓4   C2(3)   C3  o✓5  C2(3)   C3  o✓6  C2(3)  via permutations of a,b,c (these are automorphisms of C2(3)). Also C3  o✓ 1  C2(3)    C3  o✓4  C2(3)  via the automorphism (a,b,c) 7! (a,ab,c). Lastly C3 o✓ 1 C2(3)   C3 o✓7 C2(3) via the automorphism (a,b,c) 7! (a,ab,ac). Thus all seven non-trivial semi-direct products are isomorphic.

H = D8  = hy,z | y2  = z4  = 1, zy = y3 zi: as in the case of H = C4  ⇥ C2  there are three non-trivial homomorphisms: ✓ 1  = (φ2 ,id), ✓2  = (id,φ2 ), ✓3  = (φ2 , φ2 ). In this case C3  o✓2  D8   C3  o✓3  D8 . To see this let f : D8  ! D8  be defined by f(y) = yz, f(z) = z. Now ✓3 ◦ f = ✓2  and we can check that f 2 Aut(D8 ).   To see why C3 o✓ 1 D8   C3 o✓2 D8 note that ker(✓1 ) = hzi  C4 while ker(✓2 ) = hy,z2 i  (C2  ⇥ C2 ). So the former semi-direct product contains elements of order 12, such as xz, while that the later does not.

H = Q8  = hy,z | y4  = 1, y2  = z , zy = y zi23 . As in the cases of H = C4  ⇥ C2 and H = D8  there are three non-trivial homomorphisms: ✓ 1  = (φ2 ,id), ✓2  = (id,φ2 ), ✓3  = (φ2 , φ2 ). In this case they are all isomorphic. If f,g : Q8  ! Q8 f(y) = z, f(z) = y and g(y) = y, g(z) = yz, then f,g 2 Aut(Q8 ) and ✓1 ◦f = ✓2 while ✓3 ◦ g = ✓ 1 .

Overall we have identified 7 isomorphically distinct groups with n2   > 1 and n3  = 1.

We now consider n2  > 1 and n3  > 1 so n2  = 3 and n3  = 4. We will show that in this case G  S4 .

Let the 3-Sylow subgroups be S3 (G) = {P1 ,P2 ,P3 ,P4 }. Consider the action of conjugation by G on S3 (G). Sylow’s theorem tells us these subgroups are all conjugate so there is a single orbit of size 4 and so the Orbit-Stabiliser theorem tells us that for any P 2 S3 (G) the stability subgroup GP = {g 2 G | gPg−1  = P} has order 24/4 = 6.

Consider the group homomorphism associated to this action, given by φ : G ! S4   (so we interpret the action of a single group element g 2 G as a permutation of S3 (G) which in turn we view as a permutation in S4 ). If we can show that K = ker(φ) = {1} then φ is injective and hence (since |G| = 24 = |S4 |) surjective and so G  S4 . For any P 2 S3 (G) we have ker(φ) ✓ GP = {g 2 G | gPg−1  = P} so in particular for any 1 6 i < j 6 4 we have K / GPi  \ GPj .

For any 1 6 i < j 6 4 we know |GPi  \ GPj | = 1, 2, 3 or 6. If it ever has order 3 or 6 then by Sylow’s theorem it has a subgroup Q of order 3. But then Pi ,Pj are also subgroups of order 3 of GPi  \ GPj   hence, since 3-Sylow subgroups are conjugate, we have Pi  = Q = Pj . a contradiction.

Moreover if |GPi  \ GPj | = 1 then, since K is a subgroup of this group, |K| = 1 and we are done, so we may suppose that |K| = 2 and |GPi  \ GPj | = 2 for all 1 6 i < j 6 4.

Now for any P 2 S3 (G) we have K / GP  and GP  has order 6 so is either S3  or C6 . If GP   S3  then K / S3  but S3  has no normal subgroup of order 2. So we must have GP   C6 for each P 2 S3 (G). Hence each GP   C6 contains exactly two elements of order 6. Each element of order 3 or 6 belongs to a unique GP (since the intersection of any pair of stability subgroups has order 2 and so contains no elements of order 3 or 6) so G contains 2 ⇥ 4 = 8 elements of order 6 and also 2 ⇥ 4 = 8 elements of order 3. But now the remaining 8 elements of G must form a unique 2-Sylow subgroup and n2  = 1, a contradiction.

Thus |K|  2 and so |K| = 1, hence K = {1} and so φ is an isomorphism and G  S4 .

[10 marks. Full marks would have been given for either half of this answer.]

4.  (a) Prove that p(x) = x3 +9x2 +18x +13 is irreducible over Q.

Let p(x) = x  +9x  +18x+13 and note that p(x32 − 1) = x  +6x  +3x+332 . Now Eisenstein applied to s(x) = x  +6x  +3x+3 with32 p = 3 says that s(x) has no proper factorisation over Z and hence, since p(x) = s(x +1), we see that p(x) has no proper factorisation over Z. Hence, since p(x) is primitive (since it is

monic), Gauss’s Lemma then implies that r(x) is irreducible over Q. [5 marks] (b) Factorise x24 − 1 into monic irreducibles over Q, explaining carefully why each

factor is irreducible.

We use the theorem from lectures on cyclotomic polynomials: xn −1 =Qd|n Φd (x) so

x24 − 1 = Φ1 (x)Φ2 (x)Φ3 (x)Φ4 (x)Φ6 (x)Φ8 (x)Φ12 (x)Φ24 (x).

We know all of these polynomials from lectures except the Φ6 , Φ12 , Φ24 . More precisely we have Φ1 (x) = x − 1, Φ2 (x) = x +1, Φ3 (x) = x2 + x +1, Φ4 (x) = x2 +1, Φ8 (x) = x  +14 . We then find Φ6 (x)(x− 1)(x+1)(x2 +x+1) = x6 − 1 so Φ6 (x) = x2 − x+1. Next Φ12 (x)(x− 1)(x+1)(x  +x+1)(x  +1)(x222 − x+1) = x12 − 1 so Φ12 (x) = x4 − x2 + 1. Finally we can obtain Φ24  by dividing x24 − 1 by the product of the other terms: Φ24 (x) = x8 − x4 +1. So

x24 −1 = (x−1)(x+1)(x +x+1)(x +1)(x −x+1)(x +1)(x −x +1)(x −x +1)22244284 . [5 marks]

We claim these terms are all irreducible over Q. Clearly x ± 1 is irreducible since it is linear. The next three terms are all quadratics with no real roots and

hence irreducible over Q.  Finally we need to show that  Φ8 (x) , Φ 12 (x) , Φ24 (x) are irreducible over Q.  [2 marks]

For  Φ8 (x)  =  x4  + 1  we  use  Eisenstein  with p  =  2  applied  to  Φ8 (x + 1)  = x4 +4x3 +6x2 +4x +2 .  [2 marks]

A very  reasonable way  to  complete this  question  would  be to  prove that  all cyclotomic polynomials are irreducible over Q but we can also do this directly in these last two cases .

Since  Φ 12 (x)  =  x4  − x2  + 1  is  primitive  it  is  sufficient to  show  irreducibility over Z.  Now Φ 12 (x) has no integer roots (it is always at least 1 at any integer x)  so  if it  is reducible over Z then  it has  a factor x2  + ax + b.   Since  Φ 12 (x) is even either a = 0 or x2  − ax + b is a distinct factor in which case Φ 12 (x) = (x2 +ax+b)(x2 −ax+b) .  In the former case we have x4 −x2 +1 = (x2 +b)(x2 +c) and  so  bc =  1  and  b + c =  −1  with  b,c 2 Z but  this  is  impossible,  so  the latter factorisation must hold .  Equating coefficients in this case:  b2  =  1 and −1 = −a2 +2b.  So b = ±1 and a2  = −1 or a2  = 3, neither of which are possible . So  Φ 12 (x) is irreducible over Z and hence  also over Q by  Gauss’  Lemma .   [2 marks]

We can use a similar approach for Φ24 (x) = x8  − x4 + 1 .  Again Φ24 (x) has no integer roots and is even so if it is reducible over Z then it has a factor f(x) of degree at least two and at most 4 .  Let f(x) be a non-trivial factor of minimal degree .    If  deg(f)  =  3  then  f(−x)  is  a  distinct  factor  (cubic  polynomials cannot  be  even)  so  Φ24 (x)  =  f(x)f(−x)g(x)  and  deg(g)  =  2,  contradicting the minimality of deg(f) .

We would now like to assume that Φ24 (x) factors into two quartics, but we first need to consider the possibility that it factors into an even quadratic:  x2 + a and  a  sextic  x6  + bx5  + cx4  + dx3  + ex2  + fx + g,  with  a,b,c,d,e,f,g  2 Z. Equating  coefficients  we  have  b  =  d  =  f  =  0,  a + c  =  0,  e + ac  =  −1, g + ae = 0,  ag =  1 .   Now ag =  1   =) a = g =  ±1 .   If a =  1 then  c =  −1 so  e =  0  and  hence  g =  0  which  is  impossible .   Similiarly  if a =  −1 .   Thus we have a quartic factor f(x) which is either even or if not f(−x) is a distinct factor .  Thus there are two more cases:  Φ24 (x) = (x4 + ax2 + b)(x4 + cx2 + d) or Φ24 (x) = (x4 + ax3 + bx2 + cx + d)(x4 − ax3 + bx2 − cx + d) .  In the former case we have bd = 1, ad+bc = 0, c+a = 0 and −1 = b+d+ac.  Which implies a2  = −1 or a2  = 3 which is impossible .

So we have x −x +1 = (x +ax +bx +cx+d)(x −ax +bx −cx+d)84432432 .  Equating coefficients of even powers of x we have d2  = 1, 2bd = c2 , −1 = 2d − 2ac + b2 , 2b = a2 .   Since a,b,c,d 2 Z so  2b = a2  implies a is even and hence b is even but then 2d − 2ac + b2  = −1 is also even, a contradiction .

So  Φ24 (x) is irreducible over Z and hence  also over Q by  Gauss’  Lemma .   [3 marks]

(c)  Identify all possible isomorphisms between the following quotient rings:

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