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The Order of an Element


Date: 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/   
    
Associated Topics:
College Modern Algebra

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