# Isomorphism and Binary operation

## Homework Statement

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

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

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

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

## 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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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 * .

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...

Okay, so I've been trying to figure out part i, does this make sense:

If $Id_x$ is isomorphic then $Id_x (x_1 * x_2)= Id_x(x_1)* Id_x(x_2)$
$$Id_x(x_1 * x_2)= x_1 * x_2; Id_x (x_1)= x_1, Id_x(x_2)= x_2 \Rightarrow Id_x(x_1) * Id_x(x_2)= x_1 *x_2= Id_x(x_1*x_2)$$

Therefore $Id_x$ 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?

Okay, so I've been trying to figure out part i, does this make sense:

If $Id_x$ is isomorphic then $Id_x (x_1 * x_2)= Id_x(x_1)* Id_x(x_2)$
$$Id_x(x_1 * x_2)= x_1 * x_2; Id_x (x_1)= x_1, Id_x(x_2)= x_2 \Rightarrow Id_x(x_1) * Id_x(x_2)= x_1 *x_2= Id_x(x_1*x_2)$$

Therefore $Id_x$ 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 $y_1,y_2\in X_2$, that $f^{-1}(y_1*y_2)=f^{-1}(y_1)*f^{-1}(y_2)$. What does $f^{-1}(y_1),~f^{-1}(y_2),~f^{-1}(y_1*y_2)$ mean by definition?

That is correct.

Part (ii) is a bit tricky. You'll need to prove that for $y_1,y_2\in X_2$, that $f^{-1}(y_1*y_2)=f^{-1}(y_1)*f^{-1}(y_2)$. What does $f^{-1}(y_1),~f^{-1}(y_2),~f^{-1}(y_1*y_2)$ mean by definition?
Technically he just showed $Id_{x}$ is a homomorphism, no? To complete showing that it is an isomorphism, he'd have to show that $Id_{x}$ 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 $Id_{s}$ is bijective by definition" after that bit.

Technically he just showed $Id_{x}$ is a homomorphism, no? To complete showing that it is an isomorphism, he'd have to show that $Id_{x}$ 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 $Id_{s}$ is bijective by definition" after that bit.
Yes, you are correct!! It should have been added.

Technically he just showed $Id_{x}$ is a homomorphism, no? To complete showing that it is an isomorphism, he'd have to show that $Id_{x}$ 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 $Id_{s}$ 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)

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!