Find the flaw in the following argument: "Condition 2 of Theorem 5.14 is redundant, since it can be derived from 1 and 3, for let a ∈ H. Then a^-1 ∈ H by 3, and by 1, aa^-1 = e is an element of H, proving 2."

Data from 5.14 Theorem 

A subset H of a group G is a subgroup of G if and only if 

1. H is closed under the binary operation of G, 

2. the identity element e of G is in H 

3. for all a ∈ Hit is true that a^-1 ∈ H also.

Proof

The fact that if H ≤ S G then Conditions 1, 2, and 3 must hold follows at once from the definition of a sub group and from the remarks preceding Example 5.13. 

Conversely, suppose H is a subset of a group G such that Conditions 1, 2, and 3 hold. By 2 we have at once that G₂ is satisfied. Also G₃ is satisfied by 3. It remains to check the associative axiom G₁ But surely for all a, b, c ∈ H it is true that (ab)c = a(bc) in H, for we may actually view this as an equation in G, where the associative law holds. Hence H ≤  G.