Lagrange's theorem and element orders
For finite G and subgroup H, |G|=[G:H]|H|.
Key knowledge
For finite G and subgroup H, |G|=[G:H]|H|. Therefore |H| and the order of every element divide |G|; divisibility alone does not guarantee a subgroup of that order.
Conditions and scope
- G is finite and H has the inherited operation.
Sources · 1
References
Sources retain their recorded scope. A source link alone does not establish verification of every statement.
Evidence and review
Proposition 11.16, Theorem 11.17, Corollary 11.18 and Remark 11.21.
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Entry details
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- math-lagrange-theorem
- Revision
- university-core.1
- Structured reference record
- University core expansion