Groups Subgroups Normal subgroups Quotient groups Homomorphisms Cyclic groups Permutation groups Cayley s theorem Class equations Sylow theorems

Transcription

1 Group Theory Groups Subgroups Normal subgroups Quotient groups Homomorphisms Cyclic groups Permutation groups Cayley s theorem Class equations Sylow theorems

2 Groups Definition : A non-empty set ( G,*) is said to be a group with binary operation *, if it satisfies the following axioms 1. * is associative i.e a * (b*c) = (a*b) * c for all a,b,c in G 2. There exists e in G, such that a * e = e * a = a for all a in G, e is called identity in G 3. For a in G there exists a in G such that a * a = a * a = e where a is called inverse of a Note : If a * b = b * a for all a,b in G then G is called abelian group

3 Examples (Z, +), (Q,+), (R,+), (C,+) are abelian groups (Q,. ), (R,.), (C,.) are also abelian groups 1. G = { 1, -1} is abelian group w.r.t multiplication 2. G = {1, w, w 2 } then ( G,.) is an abelian group where (w 3 = 1, i.e complex cube root of unity) Here, identity = 1 and w -1 = w 2, (w 2 ) -1 = w 3. G = {1, w, w 2,, w n-1 } then ( G,.) is an abelian group where (w n = 1, i.e complex n th root of unity) Here, identity = 1 and w -1 = w n-1, (w 2 ) -1 = w n-2 4. G = { 2 r / r is an integer} then ( G,.) is an abelian group with identity = 1 and (2 r ) -1 = 2 -r 5. Let G = Zn = {0,1,2,,n-1} define * on G as a+b if a+b < n a* b = a+b-n if a+b n Then G is a group of residue modulo n. 6. Every group of order 5 is abelian

4 Cont G is abelian if O(G) = p 2 Quaternion gp (Q 8 ) = {±1, ±i, ±j, ±k} is non-abelian gp with i 2 = j 2 = k 2 = -1 Set of 2 x 2 matrices forms non-abelian gp under multiplication with determinant 0 Question : What is the condition on matrices so that it forms abelian group under multiplication? A gp G is abelian iff (ab) 2 = a 2 b 2 for all a,b in G Let (G,.) is group, for a, b in G, a m b m = b m a m and a n b n = b n a n with (m,n) = 1, m and n are integers then G is abelian. Let (G,.) is group, if (ab) i = a i b i for three consecutive integers i, a,b in G then G is abelian. Q. Give an example of abelian group in which every element is an inverse of itself. Ans. G = { 1, -1} is abelian group w.r.t multiplication

5 Cont. Q. Give and example of group in which every non-identity element is of infinite order Ans. (Z,+) The smallest non-abelian group is group of order 6 e.g S 3 Q : What is the smallest possible odd integer that can be order of a non abelian group?

6 Subgroups Definition : A non-empty subset H of group G is said to be subgroup of G if H itself forms a group under operation defined on G Theorem : A non-empty subset H of group G is subgroup of G iff i. For any a,b in H imply ab in H ii. For a in H imply a -1 in H If (G,+) is group then a,b in H imply a-b in H Example : G = Z, H = 2Z Theorem : A non-empty finite subset H of group G is subgroup of G iff H is closed under multiplication. Z(G) = { x in G / xg = gx for all g in G } is subgroup of G, called Centre of G. Remark : Z (G) = G iff G is abelian. N(a) = { x in G/ xa = ax } is subgroup of G called Normalizer or Centralizer of a in G

7 Cont. Question : What is centre of S 3? Ans. Since S 3 = { I, (12),(13),(23), (123), (132) } Z(S 3 ) = {I} G is a group of 2 x 2 non-singular matrices over reals then Z (G) = 2 x 2 scalar matrices of G If G is group in which (ab) 3 = a 3 b 3 and (ab) 5 = a 5 b 5 then G is abelian. Intersection of 2 subgroups is a subgroup. Whether Union of 2 subgroups is subgroup? Ans : No. Example : H 2 = { 2n / n in integer } H 3 = { 3n / n is integer } are subgroups of group (Z,+) then for 2,3 in H2 U H3 2-3 = -1 is not in H2 U H3 Union of 2 subgroups is subgroup iff one of them is contained in other.

8 Cont. Definition : Let H G, for a, b in G, a = b mod H if ab -1 in H. Ha = { ha / h in H } is called right coset of H in G. Cl (a) = { x in G / x = a mod H i.e. xa -1 in H } Thm : Ha = Cl (a) for all a in G. Q. Whether coset is a subgroup? Ans : No, coset is not necessarily a subgroup. Example : In Quaternian group (Q 8 ), H = { 1, -1} is subgroup for a = i, Ha = { i, -i } is not a subgroup of G.

9 Cont Lemma : There is bijection between any 2 right cosets of H in G Thm : Let H G then i) Ha = H and ah = H iff a in H ii) Ha = Hb iff ab -1 in H and ah = bh iff a -1 b in H Lagranges thm : If G is group of finite order and H G then o(h) / o(g). Example : G = Q 8 and H = { 1, -1, i, -i }, o (H)=4 / o(g)=8 Definition : Let G is a group and H G then Index of H in G is the no. of distinct right (left) cosets of H in G. It is denoted by i G (H) or [ G : H ] Remark : By Langranges thm, i G (H) = o(g) / o(h) Q. Is it possible for an infinite grp G to have a proper subgroup H with finite index.

10 Cont. Ans. Yes. Ex. G = (Z, +), H = {3n/n in Z} Then H G and has only three right cosets in Z i.e H, H+1, H+2 Definition : Let H G, C(H) = {x in G/xh = hx for all h in H} C(H) = centralizer of H in G N(H) = {x in G / xh = Hx} = {x in G / xhx -1 = H} called normalizer of H in G Remark : C(H) is subset of N(H) but C(H) N(H) Ex. G = {±1, ±i, ±j, ±k} H = {±1, ±i} Then N(H) = G and C(H) = {±1, ±i} Theorem : C(H) = G iff H is subset of Z(G) Theorem : H, K G then HK = {hk / h in H, k in K} G iff HK = KH Theorem : H, K G which are finite then o(hk) = o(h).o(k)/o(h n K) Remark : Let o(g) = finite = pq with p > q then G has at most one subgroup of order P. Ex. G = S 3, H G such that o(h) = 3 i.e H = A 3 Cor. If H, K G such that o(h) = m, o(k) = n, (m,n) = 1 then (H n K) = {e} Then o(hk) = o(h)o(k)

11 Cyclic groups Order of element : Let G be a group for a belongs to G, o(a) = n where n is least positive integer such that a n = e If (G, +) is a group then a n = na = 0 (identity) Cyclic group : A group G is called cyclic if there exists an element a in G such that every element of G can be expressed as power of a. - a is called generator of G i.e G = <a> Ex. 1. (Z, +) = <1> 2. G = {1, -1, i, -i} then (G,.) = < i > or < -i > Theorem : Order of cyclic group = order of its generator In above example, o(g) = o(i) = o(-i) = 4 Theorem : if o(a) = n in group G also a m = e then n / m Theorem : if G is finite abelian group then o(ab) / [o(a), o(b)] Theorem : A subgroup of cyclic group is also cyclic ex. G = {1, -1, i, -i}, H = { 1, -1} Remark : Every subgroup of cyclic group is normal

12 Cont Every cyclic group is abelian. Q. Whether the converse is true? i.e. Every abelian group is cyclic? Ans : No. Example : (Q, +) {Since, if it is cyclic then Q = < m/n > then for 1/3n in Q will be expressed as 1/3n = k.(m/n) which gives 1/3 = km for integers k, m (a contradiction)} Remark : If o(g) = n finite, then o(a)/o(g) Any element of finite group has finite order. Q. Is it possible that an element of infinite group is of finite order? Ans : Yes. Example : Let (Z,+) is group and G ={ Z + (m/p n )/ m,n are integers, n 0, p-fixed} Then G Q/Z where (Q,+) is group. Now, p n (Z + m/p n ) = Z + (m / p n )(pn) = Z + m = Z = Zero of G

13 Cont Order of (Z + m/p n ) divides p n Order of (Z + m/p n ) is p r ; r n Order of every element in G is finite Remark : if o(g) = n and a is an element in G such that a n = e then G is cyclic group generated by a Ex. G = {1, -1, i, -i } here o(g) = 4 and o(i) = 4 = o(-i) hence G is cyclic group generated by i or i Remark : Every subgroup of abelian group is normal Ex. G = {1, -1, i, -i } Q. Whether converse is true i.e if all subgroups of G are normal then G is abelian? Ans. No. Ex. Q 8 Theorem : Converse of Langrange s theorem holds in finite cyclic groups. Explanation : Let G = <a> such that o(g) = o(a) = n If m /n then we show that there exists H G with o(h) = m Since m / n implies n = mk, k is integer

14 Cont Let H is a cyclic group generated by a k then H G and o(h) = o(a k ) (a k ) m = a km = a n = e Suppose that (a k ) t = e then a kt = e then n / kt implies km / kt Hence m / t which gives o(a k ) = m Remark : If G is finite cyclic group with o(g) = n then number of distinct subgroups of G is number of distinct divisors of n and there is utmost one sub group of G of any given order Ex. G = <a> with o(g) = 30 then G has total 8 distinct subgroups of order 1,2,3,5,6,10,15 and 30 generated by <a>, <a 2 >, <a 3 >, <a 5 >,. <a 30 > respectively Theorem : Group of prime order must be cyclic and every element of G other than identity can be taken as its generator Q. How many non-trivial sub groups are there in group G with prime order? a) 1 b) 2 c) p d) 0

15 Cont A group of prime order cannot have any non-trivial subgroup Theorem : A group of finite composite order has atleast one non-trivial sub group Theorem : An infinite cyclic group has precisely two generators Ex. <Z, +> = < 1 or 1> The number of generators of finite cyclic group of order n is ø(n) Ex. G = Z 3 then G has ø(3) = 2 generators If G is group of order 10, then it must have sub group of order 5 Theorem : If G is finite group of order n and for every divisor d of n there exists unique subgroup of order d then G is cyclic. Ex. In cyclic group of order n, there exists ø(m) elements of order m for every divisor m of n and In a group G if ab = ba and (o(a), o(b)) = 1 then o(ab) = o(a)o(b) In a finite group G the number of elements of prime order p is multiple of p-1

16 Normal subgroup, homomorphism & permutation group Definition : H G is called normal subgroup of G if ah = Ha for all a in G It is also called invariant or self conjugate subgroup. Simple group : A group having no non-trivial normal subgroup is called simple group. Every p group is simple. Note : All odd composite ordered groups are non simple Remark : If H is normal in G and K G such that H C K C G then H is normal in K and if G is abelian then all its subgroups will be normal Ex. Since K = {1, -1, i, -i} is subgroup of G H = {1, -1} is normal in G H = {1, -1} is normal in K Theorem : A subgroup H of G is normal iff g -1 Hg = H for all g in G Theorem : A subgroup H of G is normal iff for all x,y in H and g in G, (gx) (gy) -1 in H A sub group of index 2 is always normal in G Ex. G = Q 8

17 Cont Remark : Group or prime order are only abelian simple groups - If H and K are normal subgroups in G with H n K = {e} then HK = KH