I have to proof what's in question. I know that $C(X) = \langle X \rangle $ is algebraic closure operator in $G$ and have draft of this proof in my head, but still don't get it 100%.
We have to use the defition of algebraic lattice as it is lattice which is complete and compactly generated. I know that only finetly generated subgroups are compact elements of lattice of closed subsets. See it, but can't write it down in proper way.
Could you help me with that or propose me literature which may help me with this proof? Thanks a lot!
Let $G$ be a group and $L$ the set of all subgroups of $G$. Then $L$ is partially ordered by $\subseteq$.
In fact, $L$ is a lattice: For $A,B\in L$, $A\vee B:=\bigcap\{\,H\in L\mid A,B\subseteq H\,\}$ is a least upper bound and $A\wedge B:=A\cap B$ a greatest lower bound.
In fact, this extends to arbitrary families, i.e., if $A_i\in L$ for $i\in I$, then $$\bigvee_{i\in I}A_i:=\bigcap\{\,H\in L\mid \forall i\in I\colon A_i\subseteq H\,\}$$ is a least upper bound and $$\bigwedge_{i\in I}A_i:=\bigcap_{i\in I}A_i$$ a greatest lower bound for the given family. Thus $L$ is a complete lattice.
Note that $$ \bigcup_{J\subseteq I,\atop|J|<\infty}\bigvee_{j\in J}A_j$$ is a subgroup of $G$ that contains each $A_i$, $i\in I$, i.e., the upper bound of an arbitrary family of lattice elements is in fact the settheoretic union of all upper bounds of all finite subfamilies.
Suppose $A=\langle a_1,\ldots,a_n\rangle \in L$ is finitely generated. Then $A\subseteq \bigvee_{i\in I}A_i$ implies that each $a_j$ is contained in the upper bound $A_{j,1}\vee\ldots \vee A_{j,n_j}$ over some finite subfamily. It follows that $$A=\bigvee_{j=1}^n\bigvee_{i=1}^{n_j}A_{j,i}.$$ Hence every finitely generated subgroup is a compact element of $L$.
For every $A\in L$, we have $A=\bigvee_{a\in A}\langle a\rangle$, and as each $\langle a\rangle$ is finitely generated and hence compact, we conclude that $L$ is algebraic.