[Math] If G is an abelian group and $n \in \Bbb N$, show that $\phi :G \rightarrow G$ defined by $ g \mapsto g^n$ is a group homomorphism

abstract-algebra

Quoting "If G is an abelian group and $n \in \Bbb N$, show that $\phi :G \rightarrow G$ defined by $ g \mapsto g^n$ is a group homomorphism."

My take here is that

  1. We cannot establish directly that $\phi (a \circ b) = (a \circ b)^n = a^n . b^n = \phi(a) . \phi(b)$ is necessarily true as we do not know the group operation and the affiliated set.
  2. As $g$ is mapped to $g^n$, it follows that $\phi$ is a morphism from G to the cyclic group $( \Bbb Z_n , +)$ with the following mapping:

$$\phi : g \mapsto \{1,g, …,g^{n-1} \} $$

  1. Also $Ker (\phi) =\{ g \in G:\phi(g) = 1 \}$= $\{1^n , n\in \Bbb N \}$ is a normal subgroup as G is defined as abelian: $ 1^n \circ g = g \circ 1^n$.

  2. Let $H = Ker(\phi)$ be a normal subgroup. [Not too certain about what to do here]

Any input in general and input about how to proceed is much appreciated.

Best Answer

  1. Yes you can. $G$ is a general group, you don't need to know what the language is representing, just what the rules of the game are, and the rules of the game are the group axioms. Since $G$ is abelian, you know that $ab=ba$ for all $a,b\in G$. Therefore $$\phi(ab)=(ab)^n=(ab)\cdots (ab) = (a\cdots a)(b\cdots b) = a^n b^n = \phi(a)\phi(b)$$ where the third equality follows from the commutativity of the multiplication. This is enough to prove that it is a group homomorphism.

  2. $G$ is mapped to itself. The multiplication of the group is a map $G\times G \to G$. I think you are confused with the map $\varphi_g : \mathbb{Z} \to G$, $n \mapsto g^n$.

  3. Review the definition of the kernel. in this case, the kernel is the elements with order $n$.

Related Question