Revision as of 03:53, 13 October 2015 edit 2607:f720:f00:4058:24f9:6203:1e44:9b6d (talk) No mention of requiring the identity element to be a member of the subgroup. ← Previous edit		Revision as of 04:02, 13 October 2015 edit undo 2607:f720:f00:4058:24f9:6203:1e44:9b6d (talk) Undo last change - the identity in G is actually covered later in the proof. Next edit →
Line 4: Let <math>G </math> be a group and let <math>H</math> be a nonempty subset of ~~<math>G</math>, where <math>H</math> contains the identity element from~~ <math>G</math>. If for all <math>a</math> and <math>b</math> in <math>H</math>, <math>a b ^{-1}</math> is in <math>H</math>, then <math>H</math> is a subgroup of <math>G</math>. ===Proof===

Subgroup test: Difference between revisions