First, I will state the definition of center, $i.e$, the center of group $C_G$ is defined a sthe part of $G$ which commutes with all generators.
I want to know the procedure of breaking gauge groups. Some of the textbook related with lattice gauge theories talks too much about center group and its role of breaking gauge group.
I want to know how they are related. Explaining with some examples also helps me a lot.
I'll start with describing the best known instance of center symmetry breaking:
The center of the gauge group is significant in lattice gauge theories because the expectation value of the Polyakov loop is only invariant under a central transformation if it is zero.
Hence, although the lattice action is invariant under a central transformation, this symmetry is broken when the expectation value of the Polyakov loop is non-zero, making it an order parameter for a phase transition.
Heuristically, the Polyakov loop represents $\exp(-F/T)$, where $F$ is the free energy of a single, static quark. If it is zero, this means the free energy is infinite, implying there are no single, static quarks, i.e. the theory is confining. If it is non-zero, the theory is deconfined.
Now, to the more general question how the center symmetry arises and what it is:
In a Euclidean (lattice) gauge theory at finite temperature, the gauge field $A$ must obey periodic boundary conditions in the Euclidean time direction, i.e. $$ A(t + \beta,\vec x) = A(t,\vec x)$$ and obviously, gauge transformations $g(t,x)$ with $$ g(t + \beta,\vec x) = g(t,\vec x)$$ preserve this boundary condition and leave the gauge action invariant. However, one also can have $$ g(t + \beta,\vec x) = Zg(t,\vec x)$$ for some element $Z$ from the gauge group $\mathrm{SU}(N)$. The gauge field transforms under this such that $$ A(t + \beta,\vec x) = ZA(t,\vec x)Z^\dagger$$ must hold for the boundary conditions to hold. Hence, the gauge transformation is allowed for $Z$ from the center $\mathbb{Z}_N$ of the gauge group, since then $Z$ and $Z^\dagger$ commute with all possible $A$, and $ZZ^\dagger = 1$ gives the correct boundary condition. The gauge action is invariant under such central transformations.
However, the Polyakov loop is not. It is invariant under transformation which properly respect the boundary conditions, but not under the central ones. Hence, a non-zero expectation value for the Polyakov loop breaks this central symmetry, but not the gauge symmetry as such. The addition of dynamical quarks into the action explicitly breaks center symmetry because the boundary condition for the quarks cannot be maintained by the central transformations.
The phases $Z$ that occur here are analogous to the Aharonov-Bohm phases/non-trivial Wilson loops one occurs in topologically non-trivial spacetimes - the Euclidean theory is "like a torus" due to the periodic boundary conditions, and hence topologically non-trivial.
