- #1
PsychonautQQ
- 784
- 10
Here is part of the proof via wikipedia:
"We first prove the special case that where G is abelian, and then the general case; both proofs are by induction on n = |G|, and have as starting case n = p which is trivial because any non-identity element now has order p. Suppose first that G is abelian. Take any non-identity element a, and let H be the cyclic group it generates. If p divides |H|, then a|H|/p is an element of order p. If p does not divide |H|, then it divides the order [G:H] of the quotient group G/H, which therefore contains an element of order p by the inductive hypothesis..."
I am confused why if p does not divide |H| that it must divide the index of H in G. Can anyone help me out here?
"We first prove the special case that where G is abelian, and then the general case; both proofs are by induction on n = |G|, and have as starting case n = p which is trivial because any non-identity element now has order p. Suppose first that G is abelian. Take any non-identity element a, and let H be the cyclic group it generates. If p divides |H|, then a|H|/p is an element of order p. If p does not divide |H|, then it divides the order [G:H] of the quotient group G/H, which therefore contains an element of order p by the inductive hypothesis..."
I am confused why if p does not divide |H| that it must divide the index of H in G. Can anyone help me out here?