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Apr 17 at 17:28 history edited Qmechanic CC BY-SA 4.0
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Apr 17 at 17:04 comment added ACuriousMind I don't understand this question. The center symmetry is not what you say, but for a loop along $[0,\beta]$ in the periodic time dimension, you can have a gauge transformation with $g(\beta) = ag(0)$, where $a$ is an element of the center. The Polyakov loop is not invariant under this transformation while the gauge field $A$ itself is, as one can show just by plugging in (and any good source on Polyakov loops should) - it has nothing to do with the loop "being an element of SU(3)" but with the loop being extended in $[0,\beta]$ instead of a value at a point.
Apr 17 at 16:42 history asked Joao Vitor CC BY-SA 4.0