brooke274
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If G is a set and * is an associative binary operation and G is closed under *. For all g in G there exists a unique g' in G such that g*g'*g=g. Prove (G,*) is a group. How do you prove this? Thanks in advance x
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Physics Enemy
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(Original post by brooke274)
If G is a set and * is an associative binary operation and G is closed under *. For all g in G there exists a g' in G such that g*g'*g=g. Prove (G,*) is a group. How do you prove this? Thanks in advance x
You've got associativity and closure, only need to show an identity element exists in G and that every element has an inverse.

As (g*g')*g = g and g*(g'*g) = g, do you have an identity element in G? What do the brackets suggest about the inverse of g, g' i.e) of every element in G?
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brooke274
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(Original post by Physics Enemy)
You've got associativity and closure, only need to show an identity element exists in G and that every element has an inverse.

As (g*g')*g = g and g*(g'*g) = g, do you have an identity element in G? What do the brackets suggest about the inverse of g, g' i.e) of every element in G?
I know there exists for all g in G there exists an i such that i*g=g, but how do I prove that g*g'=h*h' and even g*g'=g'*g? It's much trickier than it looks!
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Physics Enemy
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(Original post by brooke274)
I know there exists for all g in G there exists an i such that i*g=g, but how do I prove that g*g'=h*h' and even g*g'=g'*g? It's much trickier than it looks!
Indeed i*g = g*i = g, so g*g' = g'*g = i (in G by closure)
It follows that g' is the inverse of g and vice versa
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brooke274
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(Original post by Physics Enemy)
Indeed i*g = g*i = g, hence g*g' = g'*g = i
It follows g' is inverse of g and vice versa
You don't understand, sorry I must have explained it badly. I know there is an i such that i*g=g (i=g*g') and I know there exists an e (e=g'*g) such that g*e=g but how do I prove i=e. Also I how do I know that i*h=h (obviously this is true if h=g*x but this doesn't always have to be the case). I know there exists an a (a=h*h') such that a*h=h but I don't know that a=i. Does that make sense?
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MartinGrove
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(Original post by Physics Enemy)
Indeed i*g = g*i = g, so g*g' = g'*g = i (in G by closure)
It follows that g' is the inverse of g and vice versa
You cannot assume that the identity is in G.

@OP, what happens when you raise both sides of the equation to an arbitrarily large power? Remember that we are told that G is closed under the operation - in other words, the right hand side is always an element in G.

Edit: Ignore this - it did not use the 'unique' nature of g' and hence required G to be finite.
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zetamcfc
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(Original post by brooke274)
You don't understand, sorry I must have explained it badly. I know there is an i such that i*g=g (i=g*g') and I know there exists an e (e=g'*g) such that g*e=g but how do I prove i=e. Also I how do I know that i*h=h (obviously this is true if h=g*x but this doesn't always have to be the case). I know there exists an a (a=h*h') such that a*h=h but I don't know that a=i. Does that make sense?
You have shown inverses exist and that there is an identity. You are done. The identity being unique follows as you have a group.
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brooke274
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(Original post by zetamcfc)
You have shown inverses exist and that there is an identity. You are done. The identity being unique follows as you have a group.
I don't understand how the identity being unique follows. Why does if i*g=g then i*h=h?
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MartinGrove
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(Original post by brooke274)
I know there exists for all g in G there exists an i such that i*g=g, but how do I prove that g*g'=h*h' and even g*g'=g'*g? It's much trickier than it looks!
(Original post by brooke274)
I don't understand how the identity being unique follows. Why does if i*g=g then i*h=h?
From how the question is worded, I do not think you can assume that any identity exists, let alone whether it is unique or not.

I believe you are meant to show that those things follow from the closed property of this binary operation.
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zetamcfc
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(Original post by brooke274)
I don't understand how the identity being unique follows. Why does if i*g=g then i*h=h?
Suppose I have two identities, e and f. Then e=e*f=f and we're done.
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brooke274
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(Original post by zetamcfc)
You have shown inverses exist and that there is an identity. You are done. The identity being unique follows as you have a group.
I haven't shown an identity exist. I've shown for all g there exists an i such that i*g=g. An identity i is and element such for all g, i*g=g*i=g. It's just like saying that because 'for all countries there exists a capital city' is completely different to 'there exists a capital city for all countries'.
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brooke274
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(Original post by zetamcfc)
Suppose I have two identities, e and f. Then e=e*f=f and we're done.
No I have not shown there exists a number e such that e*g=g for all g, so there is no reason why e*f=f.
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Physics Enemy
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(Original post by brooke274)
You don't understand, sorry I must have explained it badly. I know there is an i such that i*g=g (i=g*g') and I know there exists an e (e=g'*g) such that g*e=g but how do I prove i=e
Proving the identity is unique? i*g = g means i*e = e, and g*e = g means i*e = i, hence e = i

(Original post by MartinGrove)
...
Q says G is closed under *, so g*g' = g'*g = i lies in G
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brooke274
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(Original post by Physics Enemy)
Proving the identity is unique? i*g = g means i*e = e, and g*e = g means i*e = i, hence e = i
i*g=g DOESN'T mean i*e=e. Cause i is dependant on g.
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brooke274
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(Original post by MartinGrove)
From how the question is worded, I do not think you can assume that any identity exists, let alone whether it is unique or not.

I believe you are meant to show that those things follow from the closed property of this binary operation.
Yes thank you. Obviously if i=g*g' then i*g=g but that doesn't mean i*h=h for all h which is what everyone is assuming. Yes obviously you've got to use the properties that * is closed and associative and create some kind of proof but unfortunetly I've been trying for ages and can't figure out why which is why I asked.

I've proved (g')'=g as g*(g'*g*g')*g=(g*g'*g)*(g'*g)=g*g'*g=g so since g*g'*g also =g and g' is unique g'*g*g'=g' so but g'*(g')'*g'=g' and we know (g')' is unique so (g')'=g.
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Physics Enemy
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(Original post by brooke274)
i*g=g DOESN'T mean i*e=e. Cause i is dependant on g.
g denotes any element in the set, including i, e
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brooke274
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(Original post by Physics Enemy)
g denotes any element in the set, including i, e
No, as we've defined i to be g*g' so i is definitely dependant on that particular value of g used in the definition of i. Maybe if we call i(g)=g*g', then i(g)*g=g but we haven't proved i(g)*h=h or that i(h)*g=g.
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Physics Enemy
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(Original post by brooke274)
No, as we've defined i to be g*g' so i is definitely dependant on that particular value of g used in the definition of i. Maybe if we call i(g)=g*g', then i(g)*g=g but we haven't proved i(g)*h=h or that i(h)*g=g.
g isn't a particular element, it represents any general element in the set. See 'for all g in G' in the Q. So no need to label elements as h, j, etc.
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brooke274
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(Original post by Physics Enemy)
g isn't a particular element, it represents any general element in the set. See 'for all g in G' in the Q. So no need to label elements as h, j, etc.
I agree g represents any general element from the set, but i is a function of g so i*g=g doesn't imply i*h=h as we can replace i with g*g'. It's just like how (1/x)*x=1 for all x but that doesn't mean (1/x)*y=y for all y, what your saying is just as non-sensical as that!
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sam23478
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Physics Enemy you just got burned! Brooke is 100% right, looks like a seriously tough problem! I'm trying it now and will let you know if I make any progress!
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