Is <4Z, +> isomorphic to <5Z, +>?

[MikeDietrich]

Homework Statement

The question asks me to determine if 4Z and 5Z (with standard addition) are isomorphic and if so to give the isomorphism.

2. The attempt at a solution

What I am having difficulty with is showing a mapping that preserves the operation. IE phi(a+b) = phi(a) + phi(b).
What I have so far is:
phi(a+b) = phi(5a/4) + phi(5b/4).
phi(a+b) = 5/4(a+b) = 5/4a + 5/4b = phi(a) + phi(b)

Therefore, my conclusion is 4Z and 5Z are isomorphic and the isomorphism is phi(n) = 5/4(n).

Can someone either confirm this or point me in the right direction? I really appreciate everyone willingness to help one another! Thank you!

 

[jbunniii]

What does "5/4" mean? This is not an element of 4Z or 5Z (nor of any nZ, for that matter). It is true that they are isomorphic, but you have to be more careful defining your map.

I suggest the following approach: can you name a third infinite additive cyclic group that is isomorphic to both 4Z and 5Z? If so, then you can conclude that 4Z is isomorphic to 5Z. This should also make it easier to see exactly what the map is.

 

[MikeDietrich]

5/4(n) is the homomorphism that would map 4Z to 5Z (yes? no?). Maybe I'm confused... I did not think/realize/know the a homomorphism needed to be in the set?

 

[Hurkyl]

What does "5/4" mean? This is not an element of 4Z or 5Z (nor of any nZ, for that matter). It is true that they are isomorphic, but you have to be more careful defining your map.

I suggest the following approach: can you name a third infinite additive cyclic group that is isomorphic to both 4Z and 5Z? If so, then you can conclude that 4Z is isomorphic to 5Z. This should also make it easier to see exactly what the map is.

I assume "5/4" refers to an element of Q. Rational number multiplication by 5/4 is a well-defined map from 4Z to Q...

Can someone either confirm this or point me in the right direction? I really appreciate everyone willingness to help one another! Thank you!

Without seeing your whole argument, it's hard to tell if you're on or nearby the right track, or if you're just stumbling around nearby.

 

[jbunniii]

5/4(n) is the homomorphism that would map 4Z to 5Z (yes? no?). Maybe I'm confused... I did not think/realize/know the a homomorphism needed to be in the set?

It's OK, but maybe a sentence would be in order to state that multiplication by the rational number 5/4 does indeed map an element in 4Z to an element in 5Z. It may seem a bit pedantic, but it's always good practice to explain why the map is well-defined if there is even the slightest doubt. Multiplying an element of 4Z by some other rational, such as 3/7, would not constitute a map from 4Z to 5Z.

So once we have established that multiplication by 5/4 is in fact a valid map (which you call phi), your argument shows that phi is a homomorphism.

However, not all homomorphisms are isomorphisms. What else needs to be true?

P.S. Another approach, which is what I was getting at, is to show that Z is isomorphic to nZ for any positive integer n. Then you automatically get nZ isomorphic to mZ for any positive integers m, n. But you don't have to do it that way.

 

[MikeDietrich]

I know I also need to show injectivity and surjectivity (I assume this is what you are getting at). In my attempt at brevity I was not clear and I thank you for your patience.

 

[jbunniii]

I know I also need to show injectivity and surjectivity (I assume this is what you are getting at). In my attempt at brevity I was not clear and I thank you for your patience.

Yes, that is what I meant. Aside from the caveat I mentioned, your proof is fine so far.

So do you have a proof of injectivity and surjectivity?