A5 is simple: Difference between revisions
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===Proof using sizes of conjugacy classes=== | ===Proof using sizes of conjugacy classes=== | ||
{{further|[[element structure of alternating group:A5]]}} | |||
The conjugacy class sizes are <math>1, 12, 12, 20, 15</math>. A normal subgroup must contain the conjugacy class of size <math>1</math>, and one or more other conjugacy classes. Thus, the order of any normal subgroup must be a sum of some of these numbers, including the <math>1</math>. By [[Lagrange's theorem]], the order must also divide the order of the group. But no such sum among these numbers divides <math>60</math>, other than <math>1</math> and <math>60</math> themselves. | The conjugacy class sizes are <math>1, 12, 12, 20, 15</math>. A normal subgroup must contain the conjugacy class of size <math>1</math>, and one or more other conjugacy classes. Thus, the order of any normal subgroup must be a sum of some of these numbers, including the <math>1</math>. By [[Lagrange's theorem]], the order must also divide the order of the group. But no such sum among these numbers divides <math>60</math>, other than <math>1</math> and <math>60</math> themselves. | ||
===Proof by listing subgroups=== | ===Proof by listing subgroups=== | ||
{{further|[[subgroup structure of alternating group:A5]]}} | |||
{{fillin}} | {{fillin}} | ||
Revision as of 16:26, 23 July 2013
Statement
The alternating group of degree five (denoted ) is a simple group.
Related facts
Proof
Proof using sizes of conjugacy classes
Further information: element structure of alternating group:A5
The conjugacy class sizes are . A normal subgroup must contain the conjugacy class of size , and one or more other conjugacy classes. Thus, the order of any normal subgroup must be a sum of some of these numbers, including the . By Lagrange's theorem, the order must also divide the order of the group. But no such sum among these numbers divides , other than and themselves.
Proof by listing subgroups
Further information: subgroup structure of alternating group:A5
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