Last time we discovered that if H⊆G, then H is a subgroup only means that H is itself a group under the same operation as G. As an example, I showed that 2Z is a subgroup of Z. But if you remember, the explination was rather frustrating and I implied that there is an easier way to show that something is a subgroup. This method is one of three subgroup tests that I will now show. I'm not going to prove that they work for the sake of time (and boredom), but if someone would like me too, I can.
Theorem: One-Step Subgroup Test
Let G be a group and H be a non-empty subset of G. If ∀a,b∈G, a∙b-1∈G, then H is a subgroup of G.
Let G be a group and H be a non-empty subset of G. If ∀a,b∈G, a∙b-1∈G, then H is a subgroup of G.
Theorem: Two-Step Subgroup Test
Let G be a group and H be a non-empty subset of G. If a∙b∈G whenever a,b∈G and x-1∈G whenever x∈G, then H is a subgroup of G.
Let G be a group and H be a non-empty subset of G. If a∙b∈G whenever a,b∈G and x-1∈G whenever x∈G, then H is a subgroup of G.
I'll now demonstrate how to use both of these tests to show that 2Z is a subgroup of Z.
Example: One-Step Subgroup Test.
Consider 2Z⊆Z. It is clear that 2Z is non-empty. Choose a,b∈2Z. By the definition of 2Z, a = 2m and b = 2n for some integers m and n. Also, the inverse of b is -b = -2n. Then we have that a+(-b) = 2m-2n = 2(m-n) and since m-n is an integer, 2(m-n)=a+(-b)∈2Z. Since the choice of a and b was arbitrary, 2Z is a subgroup of Z by the one-step subgroup test.
Consider 2Z⊆Z. It is clear that 2Z is non-empty. Choose a,b∈2Z. By the definition of 2Z, a = 2m and b = 2n for some integers m and n. Also, the inverse of b is -b = -2n. Then we have that a+(-b) = 2m-2n = 2(m-n) and since m-n is an integer, 2(m-n)=a+(-b)∈2Z. Since the choice of a and b was arbitrary, 2Z is a subgroup of Z by the one-step subgroup test.
Example: Two-Step Subgroup Test
Consider 2Z⊆Z. It is clear that 2Z is non-empty. First choose a,b∈2Z. By the definition of 2Z, a = 2m and b = 2n for some integers m and n. Thus a+b = 2m+2n = 2(m+n) and since m+n is an integer, 2(m+n)=a+b∈2Z. Second, choose x∈2Z. From the definition of 2Z we have that -x∈2Z. Since the choices of a, b, and x were arbitrary, 2Z is a subgroup of Z by the two-step subgroup test.
Consider 2Z⊆Z. It is clear that 2Z is non-empty. First choose a,b∈2Z. By the definition of 2Z, a = 2m and b = 2n for some integers m and n. Thus a+b = 2m+2n = 2(m+n) and since m+n is an integer, 2(m+n)=a+b∈2Z. Second, choose x∈2Z. From the definition of 2Z we have that -x∈2Z. Since the choices of a, b, and x were arbitrary, 2Z is a subgroup of Z by the two-step subgroup test.
You'll notice that both of these proofs looks very similar and prove the same things. However, sometimes it is easier to use one over the other. You can also see where the names come from. In the two-step test one has to show both that the subgroup is closed under its operation and that inverses are contained in the subgroup. In the one-step test both of these steps are essentially done at the same time.
There is one more subgroup test that only works for finite subgroups.
Theorem: Finite Subgroup Test
Let H be a finite, non-empty subset of G. If H is closed under the operation of G, then H is a subgroup of G.
Let H be a finite, non-empty subset of G. If H is closed under the operation of G, then H is a subgroup of G.
The finite subgroup test is nice when you know that a subgroup is finite, but this isn't always something that one can know. To prove that H⊆G is a subgroup using the finite subgroup test, you show that H is finite and then you show the first step of the two-step test.
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