IS $(\mathbb{Z}_4,+) \rightarrow (\mathbb{Z}_5^{*},\cdot), n\pmod 4 \mapsto 2^n \pmod 5 $ well-defined??

Question

For the following relation $(\mathbb{Z_4},+) \rightarrow (\mathbb{Z_5^{*}},\cdot), n\bmod 4 \mapsto 2^n \bmod 5 $

1. Determine if it is well-defined (so that it is a mapping)

Can someone show me how to do it?

So i know that to know if it is well-defined I have to show it doesn't depend on the member of the class chosen to represent it. So I take

$n_1 \equiv n_2$ that is $n_1-n_2=4k$ then $2^{n_1} =2^{n_2+4k}=16^k2^{n_2}=(5n+1)2^{n_2}=5n'+2^{n_2}$

so $2^{n_1}\equiv 2^{n_2} \pmod 5$

I am not sure, if it is correct, but still if it is I am not happy with the notation, in which I have mixed $\equiv$ with variables introduced to show one quantity is a multiple of the other. Can someone rewrite it better?

Answer 1

Let's be careful. Notice that $(\mathbb Z_4, +)$ as set is finite and has four elements and those elements are not integers.

$(\mathbb Z_4, +)$ is (using the notation used in the, rather dubious, IMO, phrase "$n\pmod 4\mapsto 2^n \pmod 5$") the set $\{[0]_4,[1]_4, [2]_4,[3]_4\}$ where the element $[k]_4$ is a class of integers $\{n\in \mathbb Z: 4|(n-k)\}$ or $\{n\in \mathbb Z: n\equiv k \pmod 4\}$ or $\{k+4m|m\in \mathbb Z\}$.

So what ther are saying is the relationship maps $[n]_4\to $ the equivalence class so that if $k\in [n]_4$ then the mapped value will be $[2^k]_5$.

The BIG assumption is that for all the $k\in [n]_4$ that all the integers $2^k$ will be in the same equivalence classes $\mod 5$.

so what we must show is that if $n \equiv m\pmod 4$ then i) $2^n \equiv 2^m\pmod 5$ and that ii) $2^n\not \equiv 0 \pmod 5$.

And that is straight forward and you did it correctly.

There is one caveat we should be aware of. Suppose $n\ge 0; n \equiv k\pmod 4$ and $k < 0$ what do we mean when we so $2^k\pmod 5$. For instance what so $\frac 18 \pmod 5$ means? Well, that actually just means what class $[a]_5$ is is so that $[a]_5*8\equiv 1\pmod 5$. That is $[a]_5 = [2]_5$.

This is fine. If $k < 0$ then $2^4\equiv 1 \pmod 5$ then $2^k\equiv 2^k*(2^{4m})\equiv 2^{k+4m}\pmod 5$ and we can just pick an $m$ that makes the whole thing positive.