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COSETS & LAGRANGE’S THEOREM
DEFINITION 1: Let H be a subgroup of a group G, 
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The set Ha = h  a : h  H  is
called a RIGHT COSET OF H in G
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For any z  Z, z + 10Z = {z + 10p: p  Z}, 10Z + z = {10p + z: p  Z}

1
EXAMPLE 2: H  
1

2
2


3
, 1, 3 is a subgroup of S3
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3


(1, 2, 3) H  (1, 2, 3), 2, 3
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EXAMPLE 3: H = {1, –1} is a subgroup of ({1, –1, i, –i},
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(–i) H = {–i, i}
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But, for an abelian group, every left coset is equal to the corresponding right coset
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A coset may or may not be a subgroup of the group
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Then,
i)

x  H  xH = H
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PROOF: i) Let h  H be chosen arbitrarily
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[h1= x 1h  H as H is a subgroup of G
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[Since H is a subgroup of G
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ii) Let x  G – H
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Then  p  xH  H
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This contradicts our hypothesis
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THEOREM 1B: Let H be a subgroup of a group G
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ii)

x  G – H  Hx  H = 
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REMARK 2: The above two theorem tells that, a coset of a subgroup made by an element of
the subgroup is the subgroup itself and a coset made by an element outside the subgroup has
no elements common with the subgroup
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3


For any x  A3, xA3 = A3
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So, A3 has only two left cosets in S3, viz
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THEOREM 2A: Let H be a subgroup of a group G
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PROOF: Let a, b  G be chosen arbitrarily
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Let
us now assume that Ha  Hb  
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 x = h1a = h2b where h1, h2  H
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Now p  Ha  p = h5a, where h5  H
 p = h5 h3b = h6b  Hb [since H being a subgroup of G, h6 = h5 h3  H]
So, Ha  Hb
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……… (2)
By (1) and (2), Ha = Hb
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THEOREM 2B: Let H be a subgroup of a group G
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PROOF: Try yourself
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For a, b  G, aH = bH if and only if

a 1b  H
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Let a 1b  h  H
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Also, b = be  bH [‘e’ is identity element of G]
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Hence by THEOREM 2B, aH = bH
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Then  h1, h2  H such that ah1= bh2
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[Since H is a subgroup of G]
THEOREM 3B: Let H be a subgroup of a group G
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PROOF: Try yourself
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Then the relation  defined on G by
‘a  b’ if and only if a 1b  H ; a, b  G, is an equivalence relation
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 a  a holds
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 

x, y  G and x  y holds  x 1 y  H  x 1 y

1

 H [since H is a subgroup of G]

 y 1 x  H  y  x holds
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u, v, w  G, u  v and v  w hold  u 1v, v 1 w  H







 u 1v v 1 w  H [since H is a subgroup of G]
 u 1 w  H [by associative property]
 u  w holds
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Hence,  is an equivalence relation
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Then the relation  defined on G by
‘a  b’ if and only if ba 1  H ; a, b  G, is an equivalence relation
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THEOREM 5A: Let H be a subgroup of a group G
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e
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PROOF: Let f: Ha  Hb be defined by f(ha) = hb  h  H
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So, f is injective
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So, f is surjective
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Consequently, Ha and Hb are equipotent
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Then for any a, b  G, aH and bH are
equipotent sets i
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, there exists a bijective mapping between aH and bH
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REMARK 3: The above two theorems show that any two left (or right) cosets of a subgroup
in a group have same cardinality
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PROOF: Let G be a finite group such that o(G) = n
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We first show that any two left cosets of H in G are either identical or disjoint [TO BE
PROVED IN THE EXAM]
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, am  G, (m  n) such that a1 H , a 2 H ,
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Then aiH  ajH =  for i, j  {1, 2, …
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Each element of G is a member of exactly one of the above cosets
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, a m H form a partition of G
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i 1

Also any two left cosets of H in G have same cardinality [TO BE PROVED IN THE
EXAM]
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[e is the identity element of G]
So, for every i  {1, 2, …
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m

Therfore, o(G) = m  o(H)
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Hence the result is proved
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The number of all distinct left (or right) cosets
of H of G is called the INDEX of H in G and is denoted by [G : H]
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[S3 : A3] = 2
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[2Z : Z] = 2
(Z, +) is a subgroup of (R, +)
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REMARK 4: From Lagrange’s Theorem, if H is a subgroup of a finite group G, then,
[G : H] =

o(G)

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