0

I have the following question:

Let $(A, +, 0)$ be a finite abelian group of order $n$.

(1) Show that the map $\varphi_k : A \to A$ defined by $a \mapsto k a$ is a homomorphism for any $k \in \mathbb{Z}$.

The solution for this is pretty straightforward. That is, $\varphi_k(a+b) = k(a+b) = ka + kb = \varphi_k(a) + \varphi_k(b)$.

(2) Show $\varphi_k$ is an isomorphism $\iff \gcd(k, n) = 1$.

This is what I have a problem with. I have tried tackle this exercise countless number of times, but I have never get a foot in. Specifically, I do not really understand how this statement makes sense, i.e. I cannot figure out how $k$ and $n$ depend on each other. From what I understand, since $\varphi_k$ is a homomorphism for any $k$, then it must be the case that the bijectivity is what requires that $\gcd(k, n) = 1$. Can someone help elucidate how $n$ and $k$ are related? Any feedback is appreciated!

Gary
  • 36,640
enfield
  • 424

1 Answers1

1

If the group is of order $n$, then $na = a+a+...+a = 0$.

Assume that $\varphi_k$ is an isomorphism. Then, $\varphi_k(a) = kb$ must have only one solution, $b = a$ and $\varphi_k$ must cover all elements of $A$, that is;

$ \{a \mid a\in A\} = \{\varphi_k(a) \mid a \in A\}$

Now assume $\text{gcd}(n,k) = d > 1$. Then, all numbers $\varphi_k(a)\in A$ must satisfy $\frac{n}{d} \varphi_k(a)= \frac{k}{d}na = 0$. Then, the order of the group $\{\varphi_k(a) \mid a \in A\}$ is $\frac{n}{d} < n$, which gives a contradiction. Then, $\text{gcd}(n,k) = 1$

Let's now assume $\text{gcd}(n,k) = 1$.

Assume there exists $b\neq a$ such that $\varphi_k(a) = \varphi_k(b)$. Then, by the homeomorphism proof, $\varphi_k(a+(-b)) = 0 \Rightarrow k(a+(-b)) = 0$. Then, there exists an element of $A$ with order $k' \mid k$. But, by Lagrange's theorem (Lagrange's theorem might be an overkill here, the main result is just that the order of an element must divide the order of a group), $k'\mid n$. Then, $k' \mid \text{gcd}(n,k) = 1$ and thus $k'=1$, meaning that $a = b$ which yields a contradiction. Thus, $\varphi_k$ is injective.

It is now easy to see surjectivity as $\varphi_k: A \to A$. By the well-known result (check here: Proof), an injective function that takes a set onto itself is also surjective. Hence, $\varphi_k$ is a bijective homomorphism, which is an isomorphism.

Please don't hesitate to correct me if I've got anything wrong or ask me questions.

by24
  • 1,140