January 24, 2021 ## (Abstract Algebra 1) The Subgroup Test

a small subgroup of individuals who have bonded together within a group is called a This is a topic that many people are looking for. voteyesons.org is a channel providing useful information about learning, life, digital marketing and online courses …. it will help you have an overview and solid multi-faceted knowledge . Today, voteyesons.org would like to introduce to you (Abstract Algebra 1) The Subgroup Test. Following along are instructions in the video below:
In order to check that a subset is a subgroup normally you would have to to check all four group axioms. So that would be closure associativity identity and inverses turns out theres a little bit of an easier way in fact. We dont need to worry about associativity and we dont need to worry about the identity.
We just need to check these two closure and inverses and so ill put this now in the form of a theorem. A non empty subset h of a group g. Is a subgroup of g.
If and only if the following properties hold and that would be closure meaning that if a and b are elements of h. Then a times b. Is an element of h.
And inverses for any element. A thats an h. A inverse is an h.
So lets look at how we might prove this so here. It is again and lets do a sketch of what we would do to try and prove this so first thing to note is that in this definition. We have an if and only if that means we have to look at both the directions.
So first lets look at the forward direction. That means that we are going to assume that we have our subgroup h. And show that the properties.
Closure and inverses hold. This is actually very very easy because before assuming now that h is a subgroup of g. So h.
Is a subgroup of g. Is where were starting from and if were assuming that well we know h is a group if its a subgroup h itself as a group and if its a group then we know that its closed and we know that as inverses. So thats actually very very easy lets look at the other direction.
Though the other direction might be a little trickier.

Although its not that much trickier thats where we assume that we have a subset h. Where we have closure and inverses and show that h is a subgroup so were going to assume to begin with that h is a sub set of g. And show that h is a sub group of g.
And we need to check the group axioms well we already have closure and we already have inverses so we have to check associativity and we have to check that there is an identity element. So for associativity lets see what can we say well. We know that h is a subset of g.
And so if we have elements a b and c. That are in h. We need to show that the associative property.
Holds namely that a times b. Times. C.
Is the same thing as a times b. Times c. Well h.
Is a subset of g. So. If a b and c.
Are an h. Isnt it also true that a b and c are in g. And if a b and c.
Are in g. And g. Is a group then associativity should hold so that works for associativity now what about an identity element.
We have to show that there is an identity element in h.

Ok. So we know that we have some element say a thats an h. And from inverses up here.
We also have an a inverse thats an h and from closure. We know that a times a inverse is an h. But what is a times a inverse a times a inverse thats the identity.
So the identity must also be an h. Ok. I think were ready for a formal proof.
Ok so to do this proof again we need to look at both directions since we have an if and only if so lets look at the forward direction. This is the easier one were going to let h be a subgroup of g. Then we need to show closure and inverses well h.
Itself is a group so it must be closed and if we know h is a group then it must be true that h. Contains. The inverse of any of its elements.
So ok. That was pretty easy. Lets look at the other direction let h.
Be a non empty subset of g. And we want to show that h is a subgroup of g. Furthermore let h.
Be closed and for any element. A thats an h let. A inverse b and h.
This is just the closure and inverses conditions to show that h is a subgroup of g.

We have to check that at the operation on h. Is associative and that h. Contains.
The identity element. This would then show that we have all four group axioms. So lets look at associativity.
If a b and c are elements of h. Then since h is a subset of g. We know that a b and c are also elements of g.
But associativity holds in g. So that means. It must also hold an h.
So we have associativity what about the identity element. Well. We know that a and a inverse are both in h.
Thats from property. Two up above and h. Is closed thats property one so it must be true that a times.
A and verses an h. But a times. A inverse is the identity.
So the identity is also an h. So this shows that h is a subgroup of g. So to summarize here.
If you ever need to show that a subset of a group as a subgroup. You dont have to check all four group axioms you only need to check closure and inverses. .