The Order of an ElementDate: 11/05/98 at 01:12:01 From: David Zdravkovic Subject: Abstract Algebra Hi, I would greatly appreciate your help with the following questions: 1) Suppose that G is a group that has exactly one nontrivial proper subgroup. Prove that G is cyclic and |G|=p^2, where p is prime. 2) If |a^2| = |b^2|, prove or disprove that |a| = |b|. 3) Let G be a finite abelian group and let a be an element of maximal order in G. Show that for every element b in G, |b| divides |a|. 4) Prove that direct product of S_3 and S_4 is not isomorphic to S_6. Thank you so much in advance. Sincerely, Dave Date: 11/05/98 at 11:11:43 From: Doctor Rob Subject: Re: Abstract Algebra 1) Consider such a group G. Suppose there are elements a and b in this group of prime orders p and q, respectively. Then the cyclic groups generated by a and b are two different proper subgroups of G, a contradiction. Thus every element of G must have order a power of a single prime number p. Pick an element a in G. If |a| = p^k with k > 2, then a^p and a^(p^2) generate two different proper subgroups of G, a contradiction. Thus every element must have order either p or p^2. Suppose every element has order p. Then take a in G. If a generates G, then G will be a cyclic group of order p that has no proper subgroups, a contradiction. Thus |G| > p. Then there is an element b in G but not in the cyclic group generated by a. The cyclic subgroup of G generated by b has order p, too, and is distinct from that generated by a. This gives us two different proper subgroups of G, a contradiction. Thus G must contain an element a of order p^2. By a similar argument, G must be cyclic and generated by a. The unique proper subgroup is the cyclic subgroup or order p generated by a^p. 2) This is false. Let a be an element of order 4*n+2, and b = a^2. 3) Show that if |b| does not divide |a|, then |a*b| > |a|. 4) For one thing, S6 contains elements of order 5, and S3 x S4 does not. For another, they have different sizes, |S3 x S4| = |S3|*|S4| = 6*24 = 144, |S6| = 720. (Are you sure that the question didn't relate to S3 x S5, which does have order 720, and elements of order 5?). For another, S6 has only one proper normal subgroup A6, while S3 x S4 has several, such as S3 x {1}, {1} x S4, A3 x A4, and so on. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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