Page 62 - DMTH403_ABSTRACT_ALGEBRA

Unit 4: Lagrange's Theorem

Now, let g  G have finite order. Then the set {e, g, g , ...} is finite, since G is finite. Therefore, all Notes
2
the powers of g can’t be distinct. Therefore, gr = gs for some r > s. Then g = e and r-s  N. Thus,
r-s
the set { t  N | g = e } is non-empty. So, by the well ordering principle it has a least element. Let
t
n be the least positive integer such that g = e.
n
Then
< g > = {e, g, g , ...., g }.
2
n-1
Therefore, o(g) = o(< g >) = n.
That is, o(g) is the least positive integer n such that g = e.
n

Note If g  (G, + ), then o(g) is the least positive integer n such that ng = e.
Now suppose g  G is of infinite order. Then, for m  11. g  gn. (Because, if g = g , then
m
n
m
g m-n = e, which shows that < g > is a finite group.) We will use this fact while proving
Theorem 5.
Theorem 4: Let G be a group and g  G be of order n. Then g = e for some m  N iff n | m.
m
Proof: We will first show that gm = e ! n (m.F or this consider the set S = { r  Z | g = e }.
r
Now, n  S. Also, if a, b  S, then g = e = g . Hence, g = ga (g ) = e. Therefore, a-b  S. Thus,
b -1
a-b
b
a
S  Z.

Note So, from last unit, we see that S = nZ. Remember, n is the least positive integer
in S!

Now if g = e for some m  N, then m  S = nZ. Therefore, n / m.
m
Now let us show that n | m  g = e. Since n | m, m = nt for some t E Z. Then g = g = (g )
m
m
nt
n t
= e = e. Hence, the theorem is proved.
t
We will now use Theorem 4 to prove a result about the orders of elements in a cyclic group.
Theorem 5: Let G = < g > be a cyclic group.
(a) If g is of infinite order then gm is also of infinite order for every m  Z.
(b) If o(g) = n, then
n
o(g ) =  m  1, ..., n 1. ((n,m) is the g.c.d. of n and m.)
m

n, m 
Proof: (a) An element is of infinite order iff all its powers are distinct. We know that all the
powers of g are distinct. We have to show that all the powers of g are distinct. If possible, let
m
(g ) = (g ) , Then g = g , But then mt = mw, and hence, t = w. This shows that the powers of g m
mt
mw
m t
m w
are all distinct, and hence g is of infinite order.
m
(b) Since o(g) = n, G = {e, g, ........ g ) . < g >, being a subgroup of G, must be of finite order. Thus,
n-1
m
n
g is of finite order. Let o(g ) = t. We will show that t = .
m
m
n, m 
LOVELY PROFESSIONAL UNIVERSITY 55