Page 71 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Therefore, < (1 2) > is not normal in S .
3
In earlier unit we proved that if H I G and K  H, then K I G. That is, ‘’ is a transitive relation. But
‘  ’ is not a transitive relation. That is, if H  N and N  G, it is not necessary that H  G.

Theorem 4: Let H and K be normal subgroups of a group G. Then H  K  G:

Proof: From Theorem 4 of Unit 3, you know that H  K  G. We have to show that
g xg  H  K  x  H  K and g  G.
-1
Now, let x EH  K and g  G. Then x  H and H  G.  g xg  H.
-1
Similarly, g xg  K.  g xg  H  K
-1
-1
Thus, H  K  G.

Example: Let G be the group generated by
{ x, y | x = e, y = e, xy = y x }
-1
4
2
L e t H = < x > a n d K = < y > .
Then show that K  G, H  G and G = HK.
Solution: Note that the elements of G are, of the form x $, where i = 0, 1 and j = 0, 1, 2, 3
i
 G = {e, x, xy, xy , xy , y , y , y }
3
2
2
3
 | G : K | = 2. Thus, by Theorem 3, K  G.
Note that we can’t apply Theorem 2, since G is non-abelian (as xy = y x and y  y ).
-1
-l
Now let us see if H  G.
Consider y xy. Now y xy = xy , because y x = xy.
-1
-1
-1
2
If xy  H, then xy = e or xy = x. (Remember o(x) = 2, so that x = x.)
2
2
-1
2
Now, xy = e  y = x = x
-1
2
2
 y = xy = y x
-1
3
 y = x
4
 e = x, a contradiction.
Again xy = x  y = e, a contradiction..
2
2
-1
2
 Y xy = xy  H, and hence, H  G.
Finally, from the definition of G you see that G = HK.
The group G is of order 8 and is called the dihedral group, D . It is the group of symmetries of a
8
square, that is, its elements represent the different ways in which two copies of a square can be
placed so that one covers the other. A geometric interpretation of its generators is shown in
figure 5.1.
Take y to be a rotation of the Euclidean plane about the origin through  , and x the reflection
2
about the vertical axis.

64 LOVELY PROFESSIONAL UNIVERSITY