# Two-step subgroup test “IFF” condition

The theorem for the two-step subgroup test says:

The subset H of a group G is a subgroup IFF the binary operation of 2 ordered pairs of elements of H are in G and for each element in H, there each exists an inverse that is in H.

Going back to the definition of Subgroup:

A subset H of a group G is a subgroup if H itself is a group under the binary operation of H,

'IF' part for the two-step subgroup test:

If H is a subgroup then H itself is a group and for it to be a group it must have already satisfied the 4 properties of the Group axiom. And we are done for 'If' part.

'Only If' part for the two-step subgroup test:

If the binary operation of two ordered pairs (a,b) are in H and the inverse of a is in H for all 'a' and 'b' in H, then H is a subgroup of G.

But the requirement for a subset H to be a subgroup is more than that! In fact, we require 2 more. How does one reconcile this?

• Which 2 more? That is not made clear, but you may be thinking, for example that a group has an identity element. That follows from the two conditions. For let $a\in H$. Then the inverse $b$ of $a$ is in $H$. And therefore $ab$, the identity element, is in $H$. You have a typo at the beginning of line 3, we want the "product" of two elements of $H$ to be in $H$. – André Nicolas May 24 '15 at 6:15
• Does the subset H of group G not also require the property of associativity and identity in order to be a subgroup? – help May 24 '15 at 6:17
• Associativity is automatically inherited, since the operation on $G$ is associative, – André Nicolas May 24 '15 at 6:20
• Does this apply in general? – help May 24 '15 at 6:28
• I don't know what in general means. Some things are automatically inherited. For example, if the group $G$ is commutative, then so is $H$. Many things are not inherited. For example, if $G$ is infinite, then $H$ need not be. – André Nicolas May 24 '15 at 6:31

So the 2 step subgroup test gives you closure and inverses. Now you just need associativity and the identity element.

We may assume H is a non-empty subset since a group must have a positive order.

If $a \in H$, then $a^{-1} \in H$ by the inverse condition.

By the closure condition, $\forall a,b \in H, ab \in H$. Apply this condition to $a,a^{-1} \in H$.

This gives us $aa^{-1}=e \in H$, so H has an identity element.

Now consider $a,b,c \in H$. $(ab)c = a(bc) \in G$, so $(ab)c = a(bc) \in H$, which gives us that the operation on H is associative.

This holds because H must be a group under the same operation as G to be a subgroup.

That gives you all four conditions and you're good to go.