Page 63 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Now, g = (g ) = e  n | tm, by Theorem 4.
mt
m t
Let d = (n, m). We can then write n = n d, m = m d, where (m , n ,) = 1.
1
1
1
1
n n
Then n  d  n, m  .
1
Now, n | tm  n | tm d  n d | tm d  n | tm .
1
1
1
1
But (n, m ) = 1. Therefore, n | t. ...... (1)
1
1
Also, (g ) 1 = g 1 = g 1 = (g ) 1 = e 1 = e.
m dn
m n
n m
m
m n
1
Thus, by definition of o(gm) and Theorem 4, we have
t | n . ...... (2)
1
(1) and (2) show that
n
t  n  n, m  .
1
n
m
i.e., o(g )  .
(n, m)
12
Using this result we know that o (4) in Z is  3.
12
12, 4 
Theorem 6: Every group of prime order is cyclic.
Proof: Let G be a group of prime order p. Since p  1,  a  G such that a  e. Theorem 4, o(a) | p.
Therefore, o(a) = 1 or o(a) = p. Since a  e, o(a)  2.
Thus, o(a) = p, i.e., o(< a >) = p. So, < a >  G such that o(< a >) = b(G). Therefore, < a > = G, that is,
G is cyclic.
Using Theorems 3 and 6, we can immediately say that all the proper subgroups of a group of
order 35 are cyclic.
Now let us look at groups of composite order.

Theorem 7: If G is a finite group such that o(G) is neither 1 nor a prime, then G has nontrivial
proper subgroups.
Proof: If G is not cyclic, then any a  G, a  e, generates a proper non-trivial subgroup < a >.

Now, suppose G is cyclic, say C = < x >, where o(x) = mn (m, n  1).
Then, (x ) = x = e. Thus, by Theorem 4, o(x )  n < o(G).
m
m n
mn
Thus <x > is a proper non-trivial subgroup of G.
m
We first define the Euler phi-function, named after the Swiss mathematician Leonard Euler
(1707-1783).
Definition: We define the Euler phi-function  : N  N as follows :
(1) = I, and
(n) = number of natural numbers < n and relatively prime to n, for n 2 2.

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