Showing that cyclic groups of the same order are isomorphic

  • #1
1,462
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Homework Statement


Prove that any two cyclic groups of the same finite order are isomorphic

Homework Equations




The Attempt at a Solution


So I began by looking at the map ##\phi : \langle x \rangle \to \langle y \rangle##, where ##\phi (x^k) = y^k##. So, I went through and showed that this is indeed an isomorphism. But when I looked at the proof in the book, is said that you first must show that this map ##\phi## is well-defined. My question here is when do I know when I should show explicitly whether a map is well-defined or not? To me it seemed relatively obvious, so I wouldn't have thought to...
 
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Answers and Replies

  • #2
member 587159

Homework Statement


Prove that any two cyclic groups of the same finite order are isomorphic

Homework Equations




The Attempt at a Solution


So I began by looking at the map ##\phi : \langle x \rangle \to \langle y \rangle##, where ##\phi (x^k) = \phi (y^k)##. So, I went through and showed that this is indeed an isomorphism. But when I looked at the proof in the book, is said that you first must show that this map ##\phi## is well-defined. My question here is when do I know when I should show explicitly whether a map is well-defined or not? To me it seemed relatively obvious, so I wouldn't have thought to...

I think you made a mistake in the map.

Shouldn't it be:

##x^k \mapsto y^k##?

The problem here is that maybe ##x^k = x^l## for ##k \neq l##. In that case, you have to check that ##y^k = y^l##, or otherwise the function isn't well-defined.

This isn't as trivial as you think (well the proof is rather short but it makes use of a theorem you proved earlier).

Also, you have to check this is an isomorphism.

Is the function a homomorphism?
Is it injective?
Is it surjective?

The first two questions should be trivial, the last one follows because we have an injection from a set to another set with the same finite order.
 
  • #3
1,462
44
I think you made a mistake in the map.

Shouldn't it be:

##x^k \mapsto y^k##?

The problem here is that maybe ##x^k = x^l## for ##k \neq l##. In that case, you have to check that ##y^k = y^l##, or otherwise the function isn't well-defined.

This isn't as trivial as you think (well the proof is rather short but it makes use of a theorem you proved earlier).

Also, you have to check this is an isomorphism.

Is the function a homomorphism?
Is it injective?
Is it surjective?

The first two questions should be trivial, the last one follows because we have an injection from a set to another set with the same finite order.
So what in general distinguishes maps that need to be checked for "well-definedness" and ones that don't? How could I tell at a glance that I should question whether one is well-defined or not? Does it have something to do with an element of the domain having more than one representation?
 
  • #4
member 587159
So what in general distinguishes maps that need to be checked for "well-definedness" and ones that don't? How could I tell at a glance that I should question whether one is well-defined or not? Does it have something to do with an element of the domain having more than one representation?

You should always check whether a map is well defined. I once wrote an entire answer to this exact same question of you. Maybe you didn't read it then. Let me know if it was useful:

https://www.physicsforums.com/threads/how-to-prove-that-a-function-is-well-defined.920532/

As a rule of thumbs, always check if a map is well-defined. Certainly when dealing with stuff that has more than 1 representation (equivalence classes, fractions, ...).
 

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