# Group theory question

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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

Last edited by brooke274; 1 year ago

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#2

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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 g' in G such that g*g'*g=g. Prove (G,*) is a group. How do you prove this? Thanks in advance x

**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

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?

Last edited by Physics Enemy; 1 year ago

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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?

**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?

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#4

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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!

**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!

It follows that g' is the inverse of g and vice versa

Last edited by Physics Enemy; 1 year ago

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Indeed i*g = g*i = g, hence g*g' = g'*g = i

It follows g' is inverse of g and vice versa

**Physics Enemy**)Indeed i*g = g*i = g, hence g*g' = g'*g = i

It follows g' is inverse of g and vice versa

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#6

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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

**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

@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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#7

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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?

**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?

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You have shown inverses exist and that there is an identity. You are done. The identity being unique follows as you have a group.

**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.

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#9

**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

I don't understand how the identity being unique follows. Why does if i*g=g then i*h=h?

**brooke274**)I don't understand how the identity being unique follows. Why does if i*g=g then i*h=h?

I believe you are meant to show that those things follow from the closed property of this binary operation.

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#10

**brooke274**)

I don't understand how the identity being unique follows. Why does if i*g=g then i*h=h?

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**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.

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(Original post by

Suppose I have two identities, e and f. Then e=e*f=f and we're done.

**zetamcfc**)Suppose I have two identities, e and f. Then e=e*f=f and we're done.

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#13

(Original post by

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

**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

(Original post by

...

**MartinGrove**)...

Last edited by Physics Enemy; 1 year ago

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(Original post by

Proving the identity is unique? i*g = g means i*e = e, and g*e = g means i*e = i, hence e = i

**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

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(Original post by

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.

**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.

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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#16

(Original post by

i*g=g DOESN'T mean i*e=e. Cause i is dependant on g.

**brooke274**)i*g=g DOESN'T mean i*e=e. Cause i is dependant on g.

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(Original post by

g denotes any element in the set, including i, e

**Physics Enemy**)g denotes any element in the set, including i, e

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#18

(Original post by

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.

**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.

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(Original post by

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.

**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.

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#20

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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