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Subgroup

Chapter 6

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Definition of SubgroupDefinition : Let G be a group and H be its subset. If H is a group under the group operation of G, then we say that H is a subgroup of G

Example: G = {1, i, −1, −i } is a group with respect to the complex number multiplication.Let S1={1} then S1 is a subgroup of GLet S2={1, −1 } then S2 is also a subgroup of GLet S3 ={1, i } then S3 is not a subgroup of G, becausethat i −1 = −i is not in S3

Let S4 ={1, i , −i } then S4 is not a subgroup of G either, because that i * (−i ) = −1 is not in S4

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Criteria of SubgroupTheorem : Let G be a group and H be its subset. Then H is asubgroup if and only if both the multiplication and inverse

operations are closure inside H , which means i) for all x, y ∈ H � xy ∈ H ii) for all x∈ H � x −1 ∈ H

Necessity is obvious. Now prove sufficiency.Let x∈ H , combine i) and ii), can get

x∈ H � x −1 ∈ H � e= xx −1 ∈ H Therefore S has identity element e. The associativity is inheritedfrom G. So H is a group and a subgroup of G.

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Theorem : Let G be a group and a∈G. Define setGa = {x: x∈G and ax=xa }

Show that Ga is a subgroup of G.Proof: Let x, y∈G . Then we have ax = xa and ay = ya So

(xy)a = x(ya) = x(ay)= (xa)y= (ax)y = a(xy)

Therefore xy∈Ga

Also from ax = xa We have x −1 (ax)x −1= x −1 (xa) x −1

x −1a (xx −1) = (xx −1)a x −1

So x −1a = a x −1

Therefore x −1 ∈Ga

That proves Ga is a subgroup of G.

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Intersection of SubgroupsTheorem : Let G be a group and H1, H2 be its two subgroups. Then their intersection H1 ∩ H2 is a subgroup too.Proof: Let x, y ∈ H1 ∩ H2

Then x, y ∈ H1 and x, y ∈ H2

Because H1 is a subgroup, then xy ∈ H1 and x −1∈H1

Because H2 is a subgroup, then xy ∈ H2 and x −1∈H2

Therefore xy ∈ H1∩ H2 and x −1∈H1∩H2

Thus H1∩ H2 is closure under multiplication and inverse operations, so it is a subgroup.

Note: Their union H1 ∪ H2 may not be a subgroup.G = {1, i, −1, −i } has two subgroup H1 = {1, i} , H1 = {1, −i} But H1 ∪ H2 = {1, i, −i} is not a subgroup.

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Subgroup generated by one elementWe define that e = a0

Theorem 1: Let G be a group and a∈G . If a has order k.Then H = {e, a, a2, …, ak−1} is a subgroup of G.We see that a−1 = ak−1 , (a2)−1 = ak−2 , …,

We define that a −n = (a −1)n

Theorem2 : Let G be a group and a ∈ G. If a has no finite order Then set

H = {all integer powers of a } = {e, a, a−1, a2, a−2, …}is a subgroup of G.It is obviously that H is closure under multiplication and inverse operations

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Order of elementTheorem 3: Let G be a finite group, which means G has finitely many elements . Then any element of G has finite order.Proof: Let a∈G . Look at all positive integer powers of a

a, a2, a3 ,…, an ,…..They are all belongs to G. However, G has only finitely many elements. Therefore at least two powers are same. Suppose that

am = an (m > n)Then am (a −1)n= an (a −1)n

So am−n = eTherefore a has finite order ≤ m − n

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Cosets of SubgroupDefinition : Let G be a group, H be a subgroup and a∈G .Define set aH = { ah :h ∈ H}which is a collection of products that multiply every element ofH by a from left side. We say that aH is a left coset of H with respect to a.

Example: G = {1, i, −1, −i } H = {1, −1}Then has four cosets:

1 H = {1, −1}i H = { i , −i}(−1) H = {−1 , 1}(− i ) H = {− i , ι}

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Theorem4: Any two left cosets of H are either same or disjoint.Proof: Let a, b∈G and aH ∩ bH is not empty.

Suppose that x∈ aH ∩ bH Then x= aha and x= bhb for some ha , hb ∈ HLet y be an element in bH. Then y=bh for some h ∈ HThus y=bh = bhb (hb)

−1h= x (hb)−1h= aha(hb)

−1h ∈ aHSo bH is subset of aH. Conversely aH is subset of bH.Therefore aH=bH

Theorem5: If G is a finite group and H is its subgroup, then every left coset of H has same number of elements as H.Proof: Let a∈G and aH be the coset. Then we can see that mapping φ : h → ah ∀ h ∈ His 1−1 and onto from H to aH. Therefore that H and aH havethe same number of elements

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Definition: The total elements of a finite group G is called its order or size, denoted as |G|

Theorem 6: Let H be a subgroup of a finite group G. Then |H| is a divisor of |G|. Proof: Because every element a∈G is in left coset aH.Thus whole group G is the union of different left cosets. From the last theorem, every left coset has the same size of |H|Suppose we have k different left cosets. Then

|G| = k |H|That proves |H| is a divisor of |G|.

Example: G = {1, i, −1, −i } cannot have subgroup of size 3,Because 3 is not divisor of 4

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Theorem 7: Let G be a finite group. The the order of any element of G is a divisor of |G|. Proof: Let a∈G . Suppose that a has order k. Then set

H = {e, a, a2, …, a k−1}is a subgroup of G and |H| = kFrom the last theorem, k is a divisor of |G|.

Theorem 8: Let G be a finite group and a∈G . If order of aequals the size of G , then G is generated by a.Proof: Suppose that a has order k. Then subgroup

H = {e, a, a2, …, a k−1}has the same size of G. So H = G

Definition: Group G is called cyclic if generated by one element,

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Theorem 9: Let G be a finite group. If size of G is a prime number, then G is cyclic.Proof: Let a∈G and a ≠ e. Then from theorem 7, order of a is a divisor of |G|. Because |G| is a prime number, its divisor is either1 or |G|. Also order(a) ≠ 1 because a ≠ e.So only possibly order(a) = |G| .Then from theorem 8, G is generated by a and G is cyclic

Example: If group G has 5 elements, then G must be generated by one element of order 5.Such as groups

Z5= { 0, 1, 2, 3, 4}2 4 6 85 5 5 51, , , ,

i i i i

G e e e eπ π π π� �

= � �� �

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Example: Find all additional subgroups of Z10

Z10 ={ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}Solution: Se of Z10 is |Z10 | = 10All possible divisor of 10 is 1, 10, 2, 5Group size is 1, the subgroup is {0}Group size is 2, the subgroup is {0, 5}Group size is 5, the subgroup is {0, 2, 4, 6, 8}Group size is 10, the subgroup is { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

Example: Find all multiplicative subgroups of Z10

Solution: Because 2×5=0, 4×5=0, 6×5=0, 8×5=0, The largest multiplicative subgroup of Z10 is G={1, 3, 7, 9}|G| =4, all possible divisors are 1, 2 , 492=1(mod 10) order(9)=2; 34 =1(mod 10) order(3)=4Subgroups are {1} { 1, 9}, {1, 3, 7, 9 }

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TheoremsTheorem 10. Let G be a group. If every non-identity element has order 2, show that G is Abelean, means G is commutative.Proof: Let a, b∈G and a, b are not identity element.Then a2 = b2 = eBecause ab has order 2 too. So

(ab)2 = ewhich means abab = eNow a(abab)b = a(e)b Therefore (aa)ba(bb) = ab And (e)ba(e) = ab So ba = ab G is commutative.

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Theorem 11. Let G be a group of size 6. Show that G must have element of order 3Proof: If one element a has order 6, then a2 has order 3.Assume that no element of G has order 3, then every non-identity element must have order 2. Let G has such two elements a, b. Let set H={e, a, b, ab}From last theorem, H is commutative, so

a2 = b2 = (ab)2 = e ∈Hba=ab∈H. a(ab)=b∈H, b(ab)=a∈H

H is closure under multiplicationAlso a −1=a ∈H, b −1=b ∈H (ab) −1=ab ∈HH is closure under inverse operation. So H is a subgroup. However, |H|=4 is not a divisor of |G|=6. This contradiction proves that G must have element of order 3

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Home works

1. Find all additional subgroups of Z11

Z11 ={ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}2. Find all multiplicative subgroups of Z11

Z11 ={ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}