Proofs on the Order of Group ElementsDate: 10/19/98 at 20:40:04 From: Jennifer V.D. Kopek Subject: Order of group elements Hi, I need help getting started on these three problems. Let a be any element of finite order of a group G. Prove: 1. If a^p = e where p is a prime number, then a has order p. (a doesn't equal e) 2. The order of a^k is a divisor of the order of a. I got this far: let ord(a) = n. Then (a^k)^n = a^(nk) = (a^n)^k = e^k = e. 3. If ord(a) = km, then ord(a^k) = m. Any help would be great. Thanks, JVDK. Date: 10/20/98 at 19:51:29 From: Doctor Tom Subject: Re: Order of group elements Hi Jennifer, 1. Clearly the order is <= p, since the order of a group element is the smallest k such that a^k = e. Since a^p = e, either p is the order, or there is some k less than p such that a^k = e. Since p is prime, if there is such a k, then k doesn't divide p, so p = nk + m, and m is less than k. e = a^p = a^(nk+m) = a^(nk)a^m = e^na^m = a^m, so a^m = e, so k wasn't the order after all. Thus p must be the order of a. 2. If the order of a^k does not divide the order of a, suppose the order of a^k = m. a^(km) = e and m is the smallest number such that this is true. If the order of a is n, then a^n = e. If m doesn't divide n then n = qm + r, with r < m. Note: e = a^(nk) = (a^k)^(qm+r) = a^(qmk)a^r = ea^r = a^r and r < m, so our assumption that m is the smallest number such that a^m = e is false. Thus the order of a^k divides the order of a. 3. If ord(a) = km, then clearly a^km = e, so the order of a^k is less than or equal to m. If it's less, say q, then (a^k)^q = a^kq = e, and kq < km, so then the order of a is also less than km, since a^kq = e. - Doctor Tom, The Math Forum http://mathforum.org/dr.math/ |
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