\[f, \; g\]

are multiplicative functions, then \[f(nm)=f(n)f(m), \; g(mn)=g(m)g(n)\]

so\[f(mn)g(mn)=f(m)f(n)g(m)g(n)=f(m)g(m)f(m)g(n)=\]

Also if we restrict the set of functions to satisfy

\[f(n)(f(n))^{-1}=1\]

and \[f(1)=1\]

the the set of arithmetical functions is a group under multiplication.\[f(mn)+g(mn)=f(m)f(n)+g(m)g(n) \neq f(m)g(m)f(m)g(n)=\]

so the set of arithmetical functions does not form a group under addition since the sum of arithmetical functions is not an arithmetical function.