= Solution
For the first <Cayley table>, <associativity> fails: $(a\ast a)\ast b=1\ast b=b$, whereas $a\ast(a\ast b)=a\ast c=d$. It therefore \b[does not define a group].
For the second table, assign the residues $0,1,2,3,4$ to $1,a,c,b,d$, respectively. Every table entry agrees with <addition modulo an integer> modulo five. Thus this table \b[defines the cyclic group of order five], with identity $1$ and inverses $a^{-1}=d$, $c^{-1}=b$. The identification with a known <group> verifies <associativity> as well as closure and the inverse axioms; inspecting only the identity row and inverses would not suffice.
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