if p^m divides |G|, show that G has a subgroup of order p^m
T ThatPinkSock New member Joined Nov 3, 2011 Messages 1 Nov 3, 2011 #1 if p^m divides |G|, show that G has a subgroup of order p^m
D daon2 Senior Member Joined Aug 17, 2011 Messages 1,003 Nov 4, 2011 #2 The base case is Cauchy's theorem. The proof of this is wide-spread, and is even on wikipedia.
