Prove G is a Cyclic GroupDate: 02/27/2003 at 17:11:41 From: Nicholas Subject: Abstract Algebra/Group Theory Let group G be finite Abelian such that G has the property that for each positive integer n the set {x in G such that x^n = identity} has at most n elements. Prove G is a cyclic group. Date: 03/01/2003 at 08:03:25 From: Doctor Jacques Subject: Re: Abstract Algebra/Group Theory Hi Nicholas, In any finite group (say of n elements), every element x of the group verifies x^n = e (this is a consequence of Lagrange's theorem). A cyclic group is a group generated by a single element (for example a). It is the set of powers of a: {e, a, a^2, ......} along with their inverses. We use the notation <a> for such a group. If a cyclic group is finite, there is a smallest integer n such that a^n = e, and the group consists of the elements: <a> = {e, a, a^2, ..., a^(n-1)} These n elements are all different, and the size of the group is therefore n. As a^i*a^(n-i) = a^n = e we see that inverses are automatically included. The order of an element is the order of the cyclic subgroup it generates (in the above example, the order of a is n). We also know that all elements x of <a> satisfy the equation x^n = e. Sorry to repeat that if you already know it, I just wanted to be sure. Let us go back to the problem at hand. What we are given is the fact that, for every n, the equation x^n = e has at most n solutions. As G is finite, we can find at least one element of maximal order. Let a be such an element, <a> the subgroup it generates, and its order be n. If <a> = G, G is cyclic by definition and we are done. Assume, for the sake of contradiction, that <a> is not equal to G. We can therefore find an element b not in <a>. Let b be of order m. As we chose n to be the maximal order of elements of G, we have m <= n. It is known that, in an Abelian group, if two elements have orders m and n, there exists an element of order equal to: p = lcm(m,n). (If you are not familiar with that property, please write back, and I'll explain it). In this case, as n is the maximal order of an element, we must have p = n (obviously, lcm(m,n) >= n). The lcm of m and n will be equal to n if and only if m divides n. This means that n = km for some k, and: b^n = b^(km) = (b^m)^k = e^k = e. and this shows that b is another solution of the equation x^n = e. As all the n elements of <a> already satisfy that equation, and b is not among them, the equation has at least n+1 solutions, contrary to the hypothesis. Does this help? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ Date: 06/21/2003 at 03:00:15 From: William Cheung Subject: Modern Algebra - Group Theory Dear Dr. Jacques and associates, fellow mathematicians, You wrote in a reply to a group theory problem: It is known that, in an Abelian group, if two elements have orders m and n, there exists an element of order equal to: p = lcm(m,n). I managed to solve the original problem, except that I couldn't prove the omitted property related to the existence of the "lcm element." I had the gut feeling that "lcm-order element" must exist in abelian group, otherwise the original theorem would be wrong. I can see that there are two approaches to prove its existence: 1) constructive approach: for any two elements a, b in Abelian Group G of order m and n, respectively, provide the closed form formula for the "lcm-order element" I tried for many weeks to arrive at such a formula, in vain. But my best guess for the formula is, assume that gcd(m,n) = k, then gcd (m/k, n/k) = 1, therefore there exist integer i and j such that i * m + j * n = 1; the "lcm-order element" *may* probably be a^j * b^i (usually multiplicative notation). I have a problem proving that it is. 2) deductive approach: to conclude that it exists with some elegant group-theoretic arguments. Thank you for your kind reply. Date: 06/21/2003 at 08:14:48 From: Doctor Jacques Subject: Re: Modern Algebra - Group Theory Hi William, Let us assume that a has order m and b has order n. We distinguish two cases: Case 1 : gcd(m,n) = 1 --------------------- In this case, lcm(m,n) = mn. Consider the intersection H of <a> and <b>. H is a subgroup, and we claim that H = {e}. If x is an element of H, <x> is a subgroup of <a> and <b>, and so the order of x must divide m and n, by Lagrange's theorem. As gcd(m,n) = 1, x has order 1, i.e. x = e. Let now x = a*b^(-1), and let k be the order of x.. As the group is Abelian, x^(mn) = (a^m)*b^(-n) = e, and k divides mn. We can write: y^k = (a^k) * b^(-k) = e a^k = b^(-k) = c As c belongs to <a> and <b>, it belongs to H, and c = e. This means that a^k = b^k = e, and k is a multiple of m and n, and therefore a multiple of lcm(m,n) = mn. As we saw that k divides mn, this shows that k = mn, and x is the required element. Case 2 : gcd(m,n) = d > 1 ------------------------- We will factor m and n as m = m1*m2 and n = n1*n2. Let p be a prime factor of mn (i.e. a prime factor of m or n), and assume that p has exponent k1 in m and exponent k2 in n (one of these exponents may be 0). If k1 >= k2, include p^k1 in m1 and p^k2 in n2; otherwise, include p^k1 in m2 and p^k2 in n1. We can check the following: * Every prime factor of mn appears in either m1 or n1, but not both, which means that gcd(m1,n1) = 1. * m1 and n1 contain all prime factors of mn with their greatest exponent, and this shows that m1*n1 = lcm(m,n). * Every prime power of m appears in either m1 or m2, but not both, therefore m = m1*m2, and, in the same way, n = n1*n2. For example, if m = 12 = 2^2*3 and n = 18 = 2*3^2, we will have m1 = 2^2, m2 = 3 n1 = 3^2, n2 = 2 All this allows us to conclude that a^m2 and b^n2 have orders m1 and n1, respectively, with gcd(m1,n1) = 1 and lcm(m1,n1) = lcm(m,n). Can you continue from here? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ |
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