Group theory question

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

(edited 4 years ago)

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Reply 1

Physics Enemy

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?

(edited 4 years ago)

Reply 2

brooke274

Original post by Physics Enemy

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!

Reply 3

Physics Enemy

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

(edited 4 years ago)

Reply 4

brooke274

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?

Reply 5

MartinGrove

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.

Reply 6

zetamcfc

Original post by brooke274

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

Reply 7

brooke274

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?

Reply 8

MartinGrove

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.

Reply 9

zetamcfc

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.

Reply 10

brooke274

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'.

Reply 11

brooke274

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.

Reply 12

Physics Enemy

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

MartinGrove

...

Q says G is closed under *, so g*g' = g'*g = i lies in G

(edited 4 years ago)

Reply 13

brooke274

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.

Reply 14

brooke274

Original post by MartinGrove

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.

Reply 15

Physics Enemy

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

Reply 16

brooke274

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.

Reply 17

Physics Enemy

Original post by brooke274

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.

Reply 18

brooke274

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!

Reply 19

sam23478

@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!