Page 59 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes  H = Hx.
(c) Hx = Hy  Hxy = Hyy = He = H xy  H, by (b)
- l
-1
–1
Conversely, xy  H  Hxy = H = Hxy y = Hy  Hx = Hy.
-l
-1
-1
Thus, we have proved (c).
The properties listed in Theorem 1 are not only true for right cosets.

Note Along the lines of the proof of Theorem 1, we can prove that if H is a subgroup of
G and x, y  G, then
(a) x  xH.

(b) xH = H  x  H.
(c) x H = yH  x y  H.
-1

Example: Let G = Sg = {I, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)} and H be the cyclic subgroup of G
generated by (1 2 3). Obtain the left cosets of H in G.
Solution: Two cosets are

H = { I, (1 2 3). (1 3 2)) and
(1 2)H = { (1 2), (1 2) × (1 2 3), (1 2) × (1 3 2))
= { (1 2) , (2 3), (1 3)}
For the other cosets you can apply Theorem 1 to see that
(1 2)H = (2 3)H = (1 3)H and

(1 2 3)H = H = (1 3 2)H.

Example: Consider the following set of 8, 2 × 2 matrices over C, Q, = (± I, ± A , ± B, ± C},
where

 1 0 0  1  0 i  i  0 
I    , A    , B    , C    and i   1.
 0 1    1 0   i 0   0  i 

You can check that the following relations hold between the elements of Q :
8
I = I , A = B = C = –I,
2
2
2
2
AB = C = –BA, BC = A = –CB, CA = B = –AC.
Therefore, Q is a non-abelian group under matrix multiplication.
8
Show that the subgroup H = < A > has only two distinct right cosets in Q .
8
Solution: H = < A > = { I, A, A , A } = {I, A, –I, –A},
2
3
since A = I, A = A, and so on.
4
5
Therefore, HB = {B, C, –B, –C} , using the relations given above.
Using Theorem 1 (b), we see that

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