abstract algebra - Are algebraic numbers analogous to group elements with finite order?

Would you say that the "elements with finite order" in group theory are analogous to "algebraic numbers" in field theory?

I thought this is the case since requiring an algebraic number $\alpha$ to be the root of a polynomial (i.e. requiring a finite combination of terms in $\alpha$ using $+$ and $\times$ to give the identity zero) is like the two-operation equivalent of an element $g$ in group theory having finite order (where there is only one operation $\times$ and we require a term in $g$ which is required to give the identity 1).

However I don't think the analogy is quite complete because in the case of the polynomial we are allowed to also multiply powers of $\alpha$ by other elements of the field to achieve the identity.

Answer

There is a general model-theoretic notion of algebraic elements, see here.

If $L/K$ is a field extension, then $a \in L$ is algebraic over $K$ in the usual sense iff it is algebraic in the sense of model theory (applied to the structure $(L,+,*,0,1)$ and the subset of $K$).

If $G$ is a group, then $a \in G$ is algebraic over $\emptyset$ in the sense of model theory iff there is some $n \in \mathbb{Z}$ with $g^n=1$ and $\{h \in G : h^n=1\}$ is finite. Thus, if $G$ is finite, then $n=0$ works and every element is algebraic. This concept is more interesting when $G$ is infinite. Then every algebraic element is torsion. The converse probably does not hold.