- #1
jmjlt88
- 96
- 0
Show that in a finite cyclic group G of order n, writtten multiplicatively, the equation xm = e has exactly m solutions x in G for each positive integers m that divides n.
Attempt...
Proof:
Let G be a finite cyclic group of order n, and suppose m is a positive integer that divides n. Let x be an element in G. Then, x generates a cyclic subgroup of G that contains (n/m) elements since the gcd(n,m)=m [we know this by a theorem presented in the section0.
Hence,
(1) (xm)n/m= xn= e.
The positive integer m was picked arbitrarily. Thus, equation (1) will hold for each m that divides n. Hence, xm=e has exactly m solutions.
QED??
Does this hold or am I missing something?
Attempt...
Proof:
Let G be a finite cyclic group of order n, and suppose m is a positive integer that divides n. Let x be an element in G. Then, x generates a cyclic subgroup of G that contains (n/m) elements since the gcd(n,m)=m [we know this by a theorem presented in the section0.
Hence,
(1) (xm)n/m= xn= e.
The positive integer m was picked arbitrarily. Thus, equation (1) will hold for each m that divides n. Hence, xm=e has exactly m solutions.
QED??
Does this hold or am I missing something?