# Cyclic Group - Isomorphism of Non Identity Mapping

1. Jul 31, 2014

### Justabeginner

1. The problem statement, all variables and given/known data
Prove that if G is a cyclic group with more than two elements, then there always exists an isomorphism: ψ: G--> G that is not the identity mapping.

2. Relevant equations

3. The attempt at a solution
So if G is a cyclic group of prime order with n>2, then by Euler's function Phi(n)> 1.
Then (r, n) = 1 by the definition of prime, and 1 < r < n. If G= (g), then Δ: g - > g^r is a nonidentity mapping.

However, this doesn't seem like a formalized proof to me.

2. Jul 31, 2014

### pasmith

There is nothing in the statement which requires that the order of G be prime, so for that reason your proof doesn't work.

Hint: Consider the map $\phi: G \to G : g \mapsto g^{-1}$.

3. Jul 31, 2014

### Justabeginner

So if G is cyclic with order n>2, and a mapping ψ: G --> G: g --> g^-1 exists, then ψ (g) = g^-1.
To check if the mapping is a homormorphism:
ψ(g) ψ(g^-1) = (g^-1)(g) = e = ψ(e) = ψ((g)(g^-1)).

This mapping is a group homomorphism. However, this is an identity mapping, so am I supposed to take another ψ(g)?

4. Jul 31, 2014

### LCKurtz

Do you know the definition of the identity map $I: G\to G$?

5. Jul 31, 2014

### Justabeginner

It is an automorphism right? A mapping of G to G itself?

6. Jul 31, 2014

### LCKurtz

What particular mapping is it? If you are trying to show something is or is not the identity map, it would be good to know what the identity map is. How is the identity map defined?

7. Jul 31, 2014

### Justabeginner

It means that when an element in group G is mapped to an element in group H, the element in group G maps to e, the identity element present in group H.

8. Jul 31, 2014

### LCKurtz

The identity element in a group is not the same thing as an identity map on the group. The identity map on a group G is the map $I: G \to G$ given by $I(a) = a$ for all $a \in G$.

9. Jul 31, 2014

### Justabeginner

So it is a mapping from itself to itself?

10. Aug 1, 2014

### HallsofIvy

Staff Emeritus
I'm not sure that sentence makes sense! The identity mapping take each element of G to itself.

11. Aug 1, 2014

### Justabeginner

Sorry that is what I meant to say. Each element of G is taken to the same element of G.

12. Aug 1, 2014

### LCKurtz

Isn't that exactly what I told you in post #8? Now, do you have any additional thoughts about what you said in the last line of post #3?

13. Aug 2, 2014

### pasmith

That is not how to check that a map is a homomorphism. You need to check that $\psi(g)\psi(h) = \psi(gh)$ for all elements $g$ and $h$. It is a consequence of this definition that $\psi(e) = e$ and $\psi(g^{-1}) = (\psi(g))^{-1}$.

Thus here
$$\psi(g)\psi(h) = g^{-1}h^{-1} = (hg)^{-1} = \psi(hg).$$ You now need to explain why the fact that $G$ is cyclic enables you to conclude that $\psi(hg) = \psi(gh)$, because it is not true for arbitrary groups that $g \mapsto g^{-1}$ is an automorphism.

The identity map on $G$ is the "do nothing" map, $g \mapsto g$.

Unless every element of $G$ is self-inverse ($g = g^{-1}$), the map $\psi$ is not the identity map on $G$. Under what circumstances is every element of a cyclic group self-inverse?

Finally, you need to show that $\psi$ is an isomorphism, not merely a homomorphism. Thus you need to explain why $\psi$ is a bijection.

14. Aug 4, 2014

### Justabeginner

Every cyclic group is abelian, so ψ(gh) = ψ(hg). I have shown that it is well defined and then injective and surjective, therefore an isomorphism. I think I've understood the technique, now off to solve more practice problems! Thank you.

15. Aug 4, 2014

### LCKurtz

You haven't shown an argument that:
1.$\psi$ is 1-1.
2.$\psi$ is onto.
3.$\psi$ is not the identity.