Subgroup test

One-step subgroup test

Proof

• Since the operation of H is the same as the operation of G, the operation is associative since G is a group.
• Since H is not empty there exists an element x in H. If we take a = x and b = x, then ab⁻¹ = xx⁻¹ = e, where e is the identity element. Therefore e is in H.
• Let x be an element in H and we have just shown the identity element, e, is in H. Then let a = e and b = x, it follows that ab⁻¹ = ex⁻¹ = x⁻¹ in H. So the inverse of an element in H is in H.
• Finally, let x and y be elements in H, then since y is in H it follows that y⁻¹ is in H. Hence x⁻¹ = xy is in H and so H is closed under the operation.

Two-step subgroup test