Swaggoholic
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A couple of quick queries i have;

If i have a group, T defined as the set of all ordered triplets, of the form (A,B,C)
Am i correct in saying that the number of elements mod(T) in T is just 3! as there is 6 ways of ordering them??


If we have 2 permutations defined as G, and H; and were asked to use the fact that
G composed with H, (G o H)= H composed with G (H o G)
to show that (G o H)^2 = e
where e is the identity permutation.

Am i right in saying that ;
(G o H)^(-1) = (Ho G)
so
(GoH)=(HoG) => (GoH)*e=(GoH)^(-1) * e
(GoH)(GoH)=e
(G o H)^2 =e

I am hoping to obtain some clarification.
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beckysoya
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I don't really inderstand this question but go on examsolution website. It is very helpful. I use that all the time.
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joostan
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(Original post by Swaggoholic)
A couple of quick queries i have;

If i have a group, T defined as the set of all ordered triplets, of the form (A,B,C)
Am i correct in saying that the number of elements mod(T) in T is just 3! as there is 6 ways of ordering them??


If we have 2 permutations defined as G, and H; and were asked to use the fact that
G composed with H, (G o H)= H composed with G (H o G)
to show that (G o H)^2 = e
where e is the identity permutation.

Am i right in saying that ;
(G o H)^(-1) = (Ho G)
so
(GoH)=(HoG) => (GoH)*e=(GoH)^(-1) * e
(GoH)(GoH)=e
(G o H)^2 =e

I am hoping to obtain some clarification.
The set of all ordered triples of ABC does indeed have 6 elements.

As for the second question, it looks to me as though you've assumed the result in order to prove the said result.
The step:
H \circ G =(G \circ H)^{-1} seems to me to be unconvincing unless I missed something.
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Swaggoholic
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(Original post by joostan)
The set of all ordered triples of ABC does indeed have 6 elements.

As for the second question, it looks to me as though you've assumed the result in order to prove the said result.
The step:
H \circ G =(G \circ H)^{-1} seems to me to be unconvincing unless I missed something.
I agree, can you please give me some advice on answering this question.
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joostan
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(Original post by Swaggoholic)
I agree, can you please give me some advice on answering this question.
Not without further clarification, sorry
With any luck someone'll be along who'll sort you out
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Swaggoholic
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(Original post by joostan)
Not without further clarification, sorry
With any luck someone'll be along who'll sort you out

thanks anyway bud
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0x2a
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The second part of your question isn't true. Take for example G = (1234), and H = (1234)(1234), where G,H \in S_4. Then G \circ H = H \circ G. However, (G \circ H)^2 = ((1234)^3)^2 = (1234)^6 = (1234)^4(1234)^2 = (1234)^2 = (13)(24) \neq 1 .
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Swaggoholic
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(Original post by 0x2a)
The second part of your question isn't true. Take for example G = (1234), and H = (1234)(1234), where G,H \in S_4. Then G \circ H = H \circ G. However, (G \circ H)^2 = ((1234)^3)^2 = (1234)^6 = (1234)^4(1234)^2 = (1234)^2 = (13)(24) \neq 1 .

The question states that,

G o H = H o G

im asked to derive that;
(G o H) ^2 = e
where e is the identity

If the questin is wrong as you suggest, can you analogue a question that works, where you can show me the method of tackling this type of question?
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0x2a
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(Original post by Swaggoholic)
The question states that,

G o H = H o G

im asked to derive that;
(G o H) ^2 = e
where e is the identity

If the questin is wrong as you suggest, can you analogue a question that works, where you can show me the method of tackling this type of question?
Where are you getting this question from? If it's from some book, could you also type in all the information that comes with it?

If the exercise is from any respectable source then they would have told you that G,H \in S_2, or that if G,H \in S_n, where n \geq 3, it should have given you some restrictions on the types of permutations that are allowed, since if G = e \Rightarrow G \circ H = H , and H^2 is not always the identity permutation.
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Swaggoholic
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(Original post by 0x2a)
Where are you getting this question from? If it's from some book, could you also type in all the information that comes with it?

If the exercise is from any respectable source then they would have told you that G,H \in S_2, or that if G,H \in S_n, where n \geq 3, it should have given you some restrictions on the types of permutations that are allowed, since if G = e \Rightarrow G \circ H = H , and H^2 is not always the identity permutation.
We define S as the set of all ordered triplets of the form (A,B,C).
G and H belong to S.

Does that help?
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james22
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(Original post by Swaggoholic)
We define S as the set of all ordered triplets of the form (A,B,C).
G and H belong to S.

Does that help?
The question is very flawed since what you are trying to prove is not even close to true. Where is the question from, and what is the FULL question. Please include all previous parts and exact wording.
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davros
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(Original post by Swaggoholic)
We define S as the set of all ordered triplets of the form (A,B,C).
G and H belong to S.

Does that help?
Not really, If S is as you describe it then G and H are just ordered triplets. So what's your law of composition for the 2 elements?
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0x2a
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(Original post by Swaggoholic)
We define S as the set of all ordered triplets of the form (A,B,C).
G and H belong to S.

Does that help?
Are you saying that S is the set of all permutations of the letters A, B and C? In that case S = S_3, and even then I don't think your claim holds ground, as (123) and (123) clearly commute, but ((123)(123))^2 = (132)^2 = (123) \neq 1.
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