Page 53 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes
Example: Consider the set K = {e, a, b, ab] and the binary operation on K given by the
4
4
table.
× e a b ab
e e a b ab
a a e ab b
b b ab e a
ab ab b a e

The table shows that (K , .) is a group.
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This group is called the Klein 4-group, after the pioneering German group theorist Felix Klein.

Example: Show that K is an abelian but not cyclic.
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Solution: From the table we can see that K is an abelian. If it were cyclic, it would have to be
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generated by e, a, b or ab. Now, < e > = {e}. Also, a = a, a = e, a”= a, and so on.
2
1
Therefore, < a > = { e, a }. Similarly, < b > = { e, b } and < ab > = { e, ab).
Therefore, K can’t be generated by e, a, b or ab.
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Thus, K is not cyclic.
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Theorem 8: Any subgroup of a cyclic group is cyclic.
Proof: Let G = < x > be a cyclic group and H be a subgroup.

If H = {e}, then H = < e >, and hence, H is cyclic.
Suppose H  {e}. Then 3 n  Z such that x H, n  0. Since H is a subgroup, (x ) = x  H.
-n
n
n -1
Therefore, there exists a positive integer m (i.e., n or -n) such that x  H. Thus, the set S = {t  N
m
| x  H) is not empty. By the well-ordering principle S has a least element, say k. We will show
t
that H = < x >.
k
Now, <x >  H, since x  H.
k
k
Conversely, let x be an arbitrary element in H. By the division algorithm n = mk + r where m,
n
r  Z, 0  r  k – 1. But then x = x n-mk = x . (x )  H, since x , x  H. But k is the least positive
n
k
r
k -m
n
n
k m
k
r
integer such that x  H. Therefore, x can be in H only if r = 0. And then, n = mk and x = (x ) 
< x >. Thus, H  < x >. Hence, H = < x >, that is, H is cyclic.
k
k
k
Now, Theorem 8 says that every subgroup of a cyclic group is cyclic. But the converse is not true.
That is, we can have groups whose proper subgroups are all cyclic, without the group being
cyclic.
Consider the group S , of all permutations on 3 symbols. Its proper subgroups are
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A = <I>
B = <(1 2)>
C = <(1 3)>
D = <(2 3)>
E = <(1 2 3)>.
As you can see, all these are cyclic. But, you know that S itself is not cyclic.
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