Via the Kan–Thurston theorem we have a map between K(G,1) and X inducing (co)homology isomorphisms. Now you either take the definition of group cohomology to be the cohomology of K(G,1), or as I view best, take the bar construction that shows the cohomology of K(G,1) is isomorphic to the algebraically-defined group cohomology (see Ken Brown's textbook).
Chris Gerig
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