Exactly n elements of order dividing n in a finite solvable group implies the elements form a subgroup
From Groupprops
Revision as of 19:41, 20 February 2009 by Vipul (talk | contribs) (New page: ==Statement== Suppose <math>G</math> is a fact about::finite solvable group, and <math>n</math> is a natural number dividing the order of <math>G</math>. If there are exactly <math>n<...)
Statement
Suppose is a Finite solvable group (?), and is a natural number dividing the order of . If there are exactly elements of whose power is the identity element, then these elements form a subgroup. This subgroup is clearly a normal subgroup. In fact, it is a characteristic subgroup, and even better, is a fully characteristic subgroup.
Facts used
Proof
PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]References
Textbook references
- The Theory of Groups by Marshall Hall, Jr., Page 145, Theorem 9.4.1, ^{More info}