- #1
jmjlt88
- 96
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Proposition: If G= <a> and b ϵ G, then the order of b is a factor of the order of a.
Proof:
Let G be a group generated by a. That is, G=<a>. Let b ϵ G. Since G is cyclic, the element b can be written as some power of a. That is, b=ak for some integer k. Suppose the order of a is n. Hence, an=e. We wish to show that the order of b is a factor of n. First, we shall divide n by k using the Division Algorithm. That is,
(1) n=qk+r for some integer q and 0≤r<n.
Then, we have,
(2) e=an=aqk+r=aqkar=e.
From (2), we observe that an=aqkar. Multiplying on the right by a-r, we obtain,
(3) ana-r=aqk. Then, we have,
(4) an-r=aqk=e.
Since an-r=e and 0≤r<n, we must have that r=0. With r=0, from (4), we get,
(5) an=aqk=(ak)q=bq=e.
Hence, the order of b is q and by (5), n=qk. That is, q|n and we have that the order of b is a factor of the order of a.
QED
Proof:
Let G be a group generated by a. That is, G=<a>. Let b ϵ G. Since G is cyclic, the element b can be written as some power of a. That is, b=ak for some integer k. Suppose the order of a is n. Hence, an=e. We wish to show that the order of b is a factor of n. First, we shall divide n by k using the Division Algorithm. That is,
(1) n=qk+r for some integer q and 0≤r<n.
Then, we have,
(2) e=an=aqk+r=aqkar=e.
From (2), we observe that an=aqkar. Multiplying on the right by a-r, we obtain,
(3) ana-r=aqk. Then, we have,
(4) an-r=aqk=e.
Since an-r=e and 0≤r<n, we must have that r=0. With r=0, from (4), we get,
(5) an=aqk=(ak)q=bq=e.
Hence, the order of b is q and by (5), n=qk. That is, q|n and we have that the order of b is a factor of the order of a.
QED