n-th power over different algebraic structure It is a classical result that the group $\mathbb{F}_p^{\times}$ is cyclic and that the equation $x^n \equiv a \pmod{p}$ is solvable iff $a^{(p-1)/gcd (p-1,n)} \equiv 1 \pmod{p}$. Also, we know that if the equation has a solution, then it has exactly $gcd (p-1, n)$ different of them. The proof uses the cyclicity of the group, so it can be gerenalize to any unit group of finite field.
This makes me wonder when is the equation $x^n=a$ in other group. For example, we know that $SL_2(\mathbb{Z})= \langle \begin{pmatrix} 1&1\\ 0&1 \end{pmatrix}, \begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix}\rangle$, or the ring of integer $\mathcal{O}_K$ of a number field $K$, which are both finitely generated. When is the equation $x^n=a$ solvable in these case? 
How about the situation in other algebraic structure? Is there some books or papers that I can read to learn more about these? Thanks in advance. 
 A: It seems that you're searching for conditions on $a$ and $n$, in order for existence of $x$ with $x^n = a$. I'm not sure how helpful this will be, but here are some thoughts:
For general groups, I don't know how much can be said: if $|G|$ is finite, a theorem of Frobenius states that for $a = 1$, the number of solutions $x$ is a multiple of $n$. Frobenius conjectured that if the number of solutions is equal to $n$, then these solutions form a subgroup of $G$: see this MO question (and answers) for more details.
If $G$ is abelian though, then one has that the $n^\text{th}$ power map $x \mapsto x^n$ is a group homomorphism, so in this case the possible $a$'s are just the image of this map. In particular, if $G$ is $n$-divisible, then $x^n = a$ is always solvable.
As to your specific examples: for matrix groups one can simply write down the $n^\text{th}$ power of a generic element, and thus obtain a (parametric) description of the possible $a$'s. As for $\mathcal{O}_K$ of a number field, you probably want $\mathcal{O}_K^\times$: this is a finitely generated abelian group, so the paragraph before applies.
