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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.

gegu
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    Your may be misunderstanding definitions in your first point. You do not need to know the specific group, this holds for all Abelian groups, so all you need are the axioms for Abelian groups. – Couchy Apr 18 '17 at 01:50

2 Answers2

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  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 $n$.

Santiago Arango-Piñeros
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    Small nitpick, the kernel is the elements with order dividing $n$, not just elements of order $n$. – Ben West Apr 18 '17 at 02:48
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Let's first prove that $\phi$ is indeed a homomorphism, then take look at your work.

First, note that $G$ is abelian, so $ab=ba$ for any arbitrary $a,b\in G$. Therefore, for any $k\in\mathbb{Z}$, the product $(ab)^k = a^kb^k$. If you're skeptical of this, try proving it using induction!

Therefore, using this fact, we have the following chain of equalities by using our definition of $\phi$:

$$ \phi(ab) = (ab)^n = a^nb^n = \phi(a)\phi(b).$$

This is the definition of a homomorphism, so we're done!

As for your work, I want to note that for $1.$ you never need an explicit description of a group to know some of its properties. Indeed, this is the power of group theory! Knowing how to deduce facts about general groups allows us to apply these facts to any possible pair of set and operation that fits our group.

Santana Afton
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