Excerpted from Beachy/Blair, Abstract Algebra, 2nd Ed. © 1996

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§ 7.7 Simple Groups

 
Lemma 7.7.1. If n 3, then every permutation in An can be expressed as a product of 3-cycles.

 
Theorem 7.7.2. The symmetric group Sn is not solvable for n 5.

 
Lemma 7.7.3. If n 4, then no proper normal subgroup of An contains a 3-cycle.

 
Theorem 7.7.4. The alternating group An is simple if n 5.

 
Definition 7.7.5. Let F be a field. The set of all n × n matrices with entries in F and determinant 1 is called the special linear group over F, and is denoted by SLn(F).
    The group SLn(F) modulo its center is called the projective special linear group and is denoted by PSLn(F).

 
Proposition 7.7.6. For any field F, the center of SLn(F) is the set of nonzero scalar matrices with determinant 1.

 
Example. 7.7.1. PSL2(F) S3 if |F| = 2.

 
Example. 7.7.2. PSL2(F) A4 if |F| = 3.

 
Lemma 7.7.7. Let F be any field. Then SL2(F) is generated by elements of the form

  and   .

 
Lemma 7.7.8. Let F be any finite field, and let N be a normal subgroup of SL2(F). If N contains an element of the form with a 0, then N = SL2(F).

 
Theorem 7.7.9. Let F be any finite field with |F| > 3. Then the projective special linear group PSL2(F) is a simple group.



§ 7.7 Simple Groups: Solved problems

 
15. Prove that there are no simple groups of order 200.     Solution

16. Sharpen Exercise 7.7.3 (b) of the text by showing that if G is a simple group that contains a subgroup of index n, where n > 2, then G can be embedded in the alternating group An.     Solution

17. Prove that if G contains a nontrivial subgroup of index 3, then G is not simple.     Solution

18. Prove that there are no simple groups of order 96.     Solution

19. Prove that there are no simple groups of order 132.     Solution

20. Prove that there are no simple groups of order 160.     Solution

21. Prove that there are no simple groups of order 280.     Solution

22. Prove that there are no simple groups of order 1452.     Solution


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