# $G$ an abelian group, $n>1$ a fixed integer, and $\phi :G\to G$ defined by $\phi(a)=a^n$ for $a\in G$. Determine wheter $\phi$ is onto.

$G$ an abelian group, $n>1$ a fixed integer, and $\phi :G\to G$ defined by $\phi(a)=a^n$ for $a\in G$. Determine wheter $\phi$ is onto.

I think it totally depends on different situations. $\forall x\in G$,we want to determine whether $x=a^n$ has solutions. But I don't know how to discuss it onto-ness under different circumstances.

• I may be misinterpreting the question, but if it is asking whether this is true for any abelian group $G$ and any $n > 1$, the answer is no. To see this, take $G = V_{4} \cong \mathbb{Z}_{2} \times \mathbb{Z}_{2}$, the Klein four group, and $n = 2$. But this is a rather trivial example, so perhaps it is asking for something stronger. Kaj Hansen's answer nicely generalizes the issue that arises with this example. Commented May 3, 2014 at 4:33

If $G$ is finite, this is not always true. In this case, all surjective homomorphisms are also injective and vice versa. So we will show that $\phi$ cannot be injective.
$\phi$ is injective $\iff \ker(\phi) = \{e\}$. Well, what if $G$ has an element of order $n$?
Even if $G$ is infinite, there are problems. Consider the group $\mathbb{Z}$ under addition. If $n > 1$, there is no $a \in \mathbb{Z}$ such that $a^n = 1$.
It is true if $|G|$ is finite and $n$ and $|G|$ are relatively prime. Can you show that? Hint: use Bezout's Lemma... $G$ does not even have to be abelian.