0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 $H$. $\endgroup$ – André Nicolas May 24 '15 at 6:15
  • $\begingroup$ Does the subset H of group G not also require the property of associativity and identity in order to be a subgroup? $\endgroup$ – help May 24 '15 at 6:17
  • $\begingroup$ Associativity is automatically inherited, since the operation on $G$ is associative, $\endgroup$ – André Nicolas May 24 '15 at 6:20
  • $\begingroup$ Does this apply in general? $\endgroup$ – help May 24 '15 at 6:28
  • $\begingroup$ 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. $\endgroup$ – André Nicolas May 24 '15 at 6:31
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.