# Group Theory - Group Axioms - Determining a group - Please help guys!

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Thread starter 5 years ago
#1
Hi

I am in need of some help when working out a maths question in order to determine whether a set is a group using the four group axioms, that is;

G1 Closure,
G2 Identity,
G3 Inverses and,
G4 Associativity.

The set is (R , o) where x o y = x + y + e

I am having real trouble even identifying the mathematics behind each of the group axioms in order to determine whether or not this forms a group! Any help in order to determine this example question would be greatly appreciated! Many thanks all!

p.s the 'R' in the set of all real numbers and also o is the binary option.
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5 years ago
#2
(Original post by alex.kundert)
Hi

I am in need of some help when working out a maths question in order to determine whether a set is a group using the four group axioms, that is;

G1 Closure,
G2 Identity,
G3 Inverses and,
G4 Associativity.

The set is (R , o) where x o y = x + y + e

I am having real trouble even identifying the mathematics behind each of the group axioms in order to determine whether or not this forms a group! Any help in order to determine this example question would be greatly appreciated! Many thanks all!

p.s the 'R' in the set of all real numbers and also o is the binary option.
The group is (R , o), the set is R.

G1: Is x o y in R? If so, prove it.
G2: Does the identity exist such that x o e = e o x = e? Prove it.
G3: Does every element have an inverse? That is, for all x do we have x o x^(-1) = x^(-1) o x = e. Prove it.
G4: If (x o y) o z = x o (y o z)? If so, prove it.
1
Thread starter 5 years ago
#3
(Original post by Zacken)
The group is (R , o), the set is R.

G1: Is x o y in R? If so, prove it.
G2: Does the identity exist such that x o e = e o x = e? Prove it.
G3: Does every element have an inverse? That is, for all x do we have x o x^(-1) = x^(-1) o x = e. Prove it.
G4: If (x o y) o z = x o (y o z)? If so, prove it.
Oh thanks a bunch for that.

For G2, the e is different to the 'e' I have posted in the question, right? Because the e in the question is relating to the irrational number e and the test in G2 is relating to the identity element?

I think my specific question was around determining whether or not each one of these axioms holds, such that they do actually show that this set constitutes a group based on the binary option. I think I am having trouble putting the proverbial 'pen to paper' and actually providing evidence against each one of these - effectively I am not particularly sure how you actually show that this is the case.
0
5 years ago
#4
(Original post by alex.kundert)
Oh thanks a bunch for that.

For G2, the e is different to the 'e' I have posted in the question, right? Because the e in the question is relating to the irrational number e and the test in G2 is relating to the identity element?

I think my specific question was around determining whether or not each one of these axioms holds, such that they do actually show that this set constitutes a group based on the binary option. I think I am having trouble putting the proverbial 'pen to paper' and actually providing evidence against each one of these - effectively I am not particularly sure how you actually show that this is the case.
Yes, they're different. It's very common to use e for the identity of an arbitrary group.

As for proving the group properties.. well, you just do it. The defining property of an identity is that id o x = x o id = x. So, if this group has an identity, then x = x o id = x + id + e. So we must have x = x + id + e for all real x. Does a number id exist which makes this true for all x? If so - you've found your identity.

Similarly for inverses. Compute x o inv(x) in symbols - can you find what inv(x) will be?

Associativity, (G4), again just do it. Compute x o (y o z) and compute (x o y) o z. Are they the same or not?

Closure's just obvious really; adding real numbers gives real numbers .
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5 years ago
#5
(Original post by FireGarden)
Yes, they're different. It's very common to use e for the identity of an arbitrary group.

As for proving the group properties.. well, you just do it. The defining property of an identity is that id o x = x o id = x. So, if this group has an identity, then x = x o id = x + id + e. So we must have x = x + id + e for all real x. Does a number id exist which makes this true for all x? If so - you've found your identity.

Similarly for inverses. Compute x o inv(x) in symbols - can you find what inv(x) will be?

Associativity, (G4), again just do it. Compute x o (y o z) and compute (x o y) o z. Are they the same or not?

Closure's just obvious really; adding real numbers gives real numbers .
Thanks for the reply.

A question regarding Closure- can you define it as addition when the set operation is + e? Is this not some form of multiplication of elements?

For the new term introduced as d ( for identity) I am not sure what you mean when you are putting id for my specific example? I dont see how that follows?

Also for the inverse, is there a quick way to find out what this would be? I cant think as to how you would find that?

Regarding associativity - how do I determine what x , y and z are in order to compute this calculation?

I honestly really appreciate all of your help guys as I am self-teaching and finding group theory in paticular a big struggle!! I miss integration lol.
Posted from TSR Mobile
0
5 years ago
#6
Let's work our way through the list:
1) Closure: this is pretty obvious: since e is a real number, and adding any three real numbers together gives you a real number (easy corollary of the fact that the real numbers form a group under addition), you're done.
2) Identity: Is there some number x such that for every other number, adding x to y then adding e gives you y again? (Yes, there is: it should be pretty obvious what x should be if you just write this equation down)
3) Inverses: If we call that identity that we've just found i (or just call it what it is, once you know what that is), and given any number x, is there something that we can add to x to get to i? (Again, this should be pretty obvious, and it should be pretty obvious what the inverse should be).
4) Distributivity: This should be easy to prove from the same for the real numbers under ordinary addition.
0
5 years ago
#7
(Original post by alex7892)
Thanks for the reply.

A question regarding Closure- can you define it as addition when the set operation is + e? Is this not some form of multiplication of elements?

For the new term introduced as d ( for identity) I am not sure what you mean when you are putting id for my specific example? I dont see how that follows?

Also for the inverse, is there a quick way to find out what this would be? I cant think as to how you would find that?

Regarding associativity - how do I determine what x , y and z are in order to compute this calculation?

I honestly really appreciate all of your help guys as I am self-teaching and finding group theory in paticular a big struggle!! I miss integration lol.
Posted from TSR Mobile
For closure, we know adding real numbers gives a real number. x o y, the operation of the group, evaluates as x+y+e, which is the addition of real numbers. So the group operation takes two real numbers, and gives out a real number. It's closed.

Suppose d is the identity element. To prove it exists it is enough to just find out what it is. If d exists, then x = x o d = x + d + e.

So then x = x + d + e, and so further 0 = d + e. Obviously then, d = -e, and that's the identity.

Inverses are done the same as the identity. Let x be any element, and write X for its inverse. Then x o X = x + X + e = -e. The first equality is how the operation works, the second is because x and X are inverse to each other - their product has to be the identity, which we found earlier to be -e. Just rearranging we find that X = -(x+2e). So there's inverses.

Now associativity. You don't need to know anything other than the group operation to do this for this group, since we know it's about real numbers.

x o ( y o z) = x o (y+z+e) = x+y+z+e+e = x+y+z+2e.

(x o y) o z = (x+y+e) o z = x+y+e+z+e = x+y+z+2e.

They match; x, y, z were arbitrary so it matches always. Thus associative.

Perhaps now with this example it's more clear what is meant by just checking the axioms.
0
5 years ago
#8
(Original post by FireGarden)
Inverses are done the same as the identity. Let x be any element, and write X for its inverse. Then x o X = x + X + e = -e. The first equality is how the operation works, the second is because x and X are inverse to each other - their product has to be the identity, which we found earlier to be -e. Just rearranging we find that X = -(x+e). So there's inverses.

x o ( y o z) = x o (y+z+e) = x+y+z+e+e = x+y+z+2e.

(x o y) o z = (x+y+e) o z = x+y+e+z+e = x+y+z+2e.
You should probably have a quick note saying that addition over the reals is commutative when simplifying the associative thing into x+ y + z + 2e (like how you justified closure by using the closure of real numbers under addition).

Also, I might be going crazy here, but is it not - probably just a typo on your part, but I'm pointing it out in case the OP gets confused.
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5 years ago
#9
(Original post by Zacken)
You should probably have a quick note saying that addition over the reals is commutative when simplifying the associative thing into x+ y + z + 2e (like how you justified closure by using the closure of real numbers under addition).

Also, I might be going crazy here, but is it not - probably just a typo on your part, but I'm pointing it out in case the OP gets confused.
You're right, it's -(x+2e). Silly slip.

When I wrote the final =x+y+z+2e on both lines, I intended that to be the statement "because real number addition lets us do this (i.e., they're commutative); see that the results are the same". I guess it may be too terse, but now your post explains it .
0
5 years ago
#10
(Original post by FireGarden)
You're right, it's -(x+2e). Silly slip.

When I wrote the final =x+y+z+2e on both lines, I intended that to be the statement "because real number addition lets us do this (i.e., they're commutative); see that the results are the same". I guess it may be too terse, but now your post explains it .
Great! Hopefully this is enough for the OP. :-)
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Thread starter 5 years ago
#11
(Original post by Zacken)
Great! Hopefully this is enough for the OP. :-)
Yes, this really is. Thanks for investing your time above Zacken and FireGarden into helping me understand this concept and specific example! Really appreciated! Have a nice evening both,
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