Page 60 - DMTH403_ABSTRACT_ALGEBRA

Unit 4: Lagrange's Theorem

H = HI = HA = H(–I) = H(–A). Notes
Using Theorem 1 (c), we see that

HB = HC = H(–B) = H(–C).
Therefore, H has only two distinct right cosets in Q , H and HB.
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We will now show that each group can be written as the union of disjoint cosets of any of its
subgroups. For this, first we define a relation on the elements of G.
Definition: Let H be a subgroup of a group G. We define a relation ‘~’ on G by x – y iff x y  H,
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where x, y  G. Thus, from Theorem 1 we see that x – y iff Hx = Hy.
We will prove that this relation is an equivalence relation.
Theorem 2: Let H be a subgroup of a group G. Then the relation ~ defined by ‘x ~ y iff xy  H’
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is an equivalence relation. The equivalence classes are the right cosets of H in G.
Proof: We need to prove that ~ is reflexive, symmetric and transitive.
Firstly, for any x  G, xx = e  H.  x ~ x, that is, ~ is reflexive.
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Secondly, if x ~ y for any x, y  G, then xy  H.
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 (xy ) = yx  H. Thus, y ~ x. That is, ~ is symmetric.
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Finally, if x, y, z  G such that x ~ y and y ~ z, then xy  H and yz  H.
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 (xy ) (yz ) = x (y y)z = xz  H.  x ~ Z.
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That is , ~ is transitive.
Thus, ~ is an equivalence relation.
The equivalence class determined by x  G is
[ x l = { y  G | y ~ x } = { y  G | xy  H}.
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Now, we will show that [x] = Hx. So, let y  [X}. Then Hy = Hx, by Theorem 1. And since y  Hy,
y  Hx.
Therefore, [x]  Hx.
Now, consider any element hx of Hx. Then x(hx) = xx h = h  H.
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Therefore, hx ~ x. That is, hx  [x]. This is true for any hx  Hx. Therefore, Hx  [x].
Thus, we have shown that [x] = Hx.
Remark: If Hx and Hy are two right cosets of a subgroup H in G, then W = W or HX  HY = 
y
x
Note that what Theorem 2 and the remark above say is that any subgroup H of a group G
partitions G into disjoint right cosets.
On exactly the same lines as above we can state that:
(i) any two left cosets of H in G are identical or disjoint, and

(ii) G is the disjoint union of the distinct left cosets of H in G.
4.2 Lagrange's Theorem

To understand this theorem first we have to define the order of a finite group, after that we will
show that the order of any subgroup divides the order of the group.

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