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Isomorphism and Binary operation

  • Thread starter Gale
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Homework Statement


(i) If [itex](X,*) [/itex] is a binary operation, show that the identity function
[itex] Id_X : X \rightarrow X [/itex]is an isomorphism.

(ii) Let [itex](X_1, *_1) and (X_2, *_2)[/itex] be two binary structures and let [itex]f : X_1 \rightarrow X_2[/itex] be an isomorphism of the binary structures. Show that [itex]f^-1 : X_2 \rightarrow X_1 [/itex] is also an isomorphism.

(iii) Let [itex] (X_1, *_1), (X_2, *_2), (X_3, *_3) [/itex] be three binary structures and
let [itex] f : X_1 \rightarrow X_2 [/itex] and [itex] g : X_2 \rightarrow X_3 [/itex] be isomorphisms of the binary structures. Show that [itex] g \circ f : X_1 \rightarrow X_3 [/itex] is also an isomorphism.

(iv) Denote the statement that [itex](X_1,*_1) [/itex] and [itex] (X_2, *_2) [/itex]are isomorphic by [itex](X_1, *_1) \cong (X_2, *_2)[/itex]. Using the above, show that [itex] \cong[/itex] is reflexive, symmetric and transitive.


Homework Equations





The Attempt at a Solution



Okay, so I'm a bit confused with how to work with isomorphisms and binary operations in general. I'm don't know how to approach the first half of the problem, so I can't really do the rest either. Am I supposed to choose elements from the set X and work my proofs from there, or is there some other approach I should be taking? Besides that, I'm not sure I entirely understand the more general premise of the problems:

Starting with i) I'm not sure why the identity function is only isomorphic when there exists a binary relation. I'm not very confident in my understanding, but it seems like the identity function would always be isomorphic?

ii) I'm not sure how to start a proof for this, but since f is an isomorphism, isn't it necessarily bijective so obviously it would have an inverse? I'm confused as to what the proof is supposed to prove?
 
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Answers and Replies

  • #2
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the definition of an isomorphism is that the function is bijective and linear. I think the first question pertains to the identical function, which does not (?) make use of * .
 
  • #3
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the definition of an isomorphism is that the function is bijective and linear. I think the first question pertains to the identical function, which does not (?) make use of * .
This is why I don't understand the problem... I think. The identity function isn't ismorphic with itself, because its only operating on the set X, and isomorphisms are from functions to functions? But I suppose the question could be relating the Id function to the initial binary operation? But if that's the case, i'm still not sure where to start the proof for that...
 
  • #4
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Okay, so I've been trying to figure out part i, does this make sense:

If [itex] Id_x[/itex] is isomorphic then [itex] Id_x (x_1 * x_2)= Id_x(x_1)* Id_x(x_2)[/itex]
[tex] Id_x(x_1 * x_2)= x_1 * x_2; Id_x (x_1)= x_1, Id_x(x_2)= x_2 \Rightarrow Id_x(x_1) * Id_x(x_2)= x_1 *x_2= Id_x(x_1*x_2)[/tex]

Therefore [itex] Id_x[/itex] is isomorphic?

For part ii) I was trying to use a similar argument, but I'm not sure it works. Is there a different way to approach it?
 
  • #5
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Okay, so I've been trying to figure out part i, does this make sense:

If [itex] Id_x[/itex] is isomorphic then [itex] Id_x (x_1 * x_2)= Id_x(x_1)* Id_x(x_2)[/itex]
[tex] Id_x(x_1 * x_2)= x_1 * x_2; Id_x (x_1)= x_1, Id_x(x_2)= x_2 \Rightarrow Id_x(x_1) * Id_x(x_2)= x_1 *x_2= Id_x(x_1*x_2)[/tex]

Therefore [itex] Id_x[/itex] is isomorphic?
That is correct.

For part ii) I was trying to use a similar argument, but I'm not sure it works. Is there a different way to approach it?
Part (ii) is a bit tricky. You'll need to prove that for [itex]y_1,y_2\in X_2[/itex], that [itex]f^{-1}(y_1*y_2)=f^{-1}(y_1)*f^{-1}(y_2)[/itex]. What does [itex]f^{-1}(y_1),~f^{-1}(y_2),~f^{-1}(y_1*y_2)[/itex] mean by definition?
 
  • #6
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That is correct.



Part (ii) is a bit tricky. You'll need to prove that for [itex]y_1,y_2\in X_2[/itex], that [itex]f^{-1}(y_1*y_2)=f^{-1}(y_1)*f^{-1}(y_2)[/itex]. What does [itex]f^{-1}(y_1),~f^{-1}(y_2),~f^{-1}(y_1*y_2)[/itex] mean by definition?
Technically he just showed [itex]Id_{x}[/itex] is a homomorphism, no? To complete showing that it is an isomorphism, he'd have to show that [itex]Id_{x}[/itex] is also bijective, granted that follows rather trivially from the definition of the identity function, but I think he should still state something like "And [itex]Id_{s}[/itex] is bijective by definition" after that bit.
 
  • #7
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Technically he just showed [itex]Id_{x}[/itex] is a homomorphism, no? To complete showing that it is an isomorphism, he'd have to show that [itex]Id_{x}[/itex] is also bijective, granted that follows rather trivially from the definition of the identity function, but I think he should still state something like "And [itex]Id_{s}[/itex] is bijective by definition" after that bit.
Yes, you are correct!! It should have been added.
 
  • #8
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Technically he just showed [itex]Id_{x}[/itex] is a homomorphism, no? To complete showing that it is an isomorphism, he'd have to show that [itex]Id_{x}[/itex] is also bijective, granted that follows rather trivially from the definition of the identity function, but I think he should still state something like "And [itex]Id_{s}[/itex] is bijective by definition" after that bit.
Thanks... someone else noticed that today as well. I suppose I wasn't solid on the definition of an isomorphism... so I didn't realize I needed a bijection. We haven't spent any time discussing other morphisms, so I didn't understand the difference.

(Also, I'm a she, not he)
 
  • #9
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Thanks... someone else noticed that today as well. I suppose I wasn't solid on the definition of an isomorphism... so I didn't realize I needed a bijection. We haven't spent any time discussing other morphisms, so I didn't understand the difference.

(Also, I'm a she, not he)
English needs a pronoun for when one does not know gender xD. Using "they" sounds so awkward when referring to one person!
 

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