Let G be a group of order 29 Find the center of G ZGSolution

Let G be a group of order 29. Find the center of G, Z(G)?

29 is prime and we know every group of prime order is cyclic

=> G is a cyclic group

=> G is abelian ( since every cyclic group is abelian)

=> G= Z(G) ( since each element of G must commute with each element of G and Z(G)= { z belongs to G : zg=gz for z belongs to G}

So, center of G = Z(G) = G