First we verify that ( ∼) defines an equivalence relation on 𝐽(𝑅), where the only condition that isn't immediately obvious is transitivity. Suppose 0 ≠𝑎,𝑏,𝛽,𝛾 ∈𝑅 so that

Then 𝑎𝛽𝔞 =𝑏𝛽𝔟 =𝑏𝛾𝔠, so 𝔞 ∼𝔠, as required.

Next we show that ( ∼) defines a congruence relation on the monoid 𝐽(𝑅). Now suppose 𝔞,𝔞′,𝔟,𝔟′ ∈𝐽(𝑅) such that 𝑎𝔞 =𝛼𝔞′ and 𝑏𝔟 =𝛽𝔟′. Then 𝑎𝑏𝔞𝔟 =𝛼𝛽𝔞′𝔟′ so 𝔞𝔟 ∼𝔞′𝔟′. The quotient monoid 𝐽(𝑅)/( ∼) is thence well-defined.

Now we show that 𝐽(𝑅)/( ∼) is in fact a group. Let 𝔞 ∈𝐽(𝑅) and 0 ≠𝑎 ∈𝔞. Then ⟨𝑎⟩ ⊆𝔞 so 1 ∼⟨𝑎⟩ =𝔞𝔟 for some ideal 𝔟.

Finally we show that these groups are isomorphic, letting 𝐺 and ˜𝐺 denote the constructions with and without fraction ideals respectively. An arbitrary element in 𝐺 is 𝐴𝑃(𝑅) for some fractional ideal 𝐴. But 𝑟𝐴 =𝔞 for some 𝑟 ∈𝑅 and 𝔞 ∈𝐼(𝐽), so 𝔞𝑃(𝑅) =𝐴⟨𝑟⟩𝑃(𝑅). Therefore we can always use an integral ideal as a representative for an element of 𝐺, which itself represents an element of ˜𝐺. Clearly 𝔞𝑃(𝑅) =𝔟𝑃(𝑅) ⟺ 𝔞 ∼𝔟: hence we have an isomorphism.