Since all surjective functions are epic, it follows that surjective homomorphisms are too. The converse requires more care. Let 𝜙 :𝐺 ↠𝐻 be an epimorphism and 𝐴 =𝜙(𝐺). It is sufficient to show 𝐴 =𝐻. Consider the set 𝐺/𝐴 of left-cosets of 𝐴 in 𝐺, and let 𝑢 be a unique symbol, 𝑆 =𝐺/𝐴 ⨿{𝑢}, and 𝑆! denote the permutation group of the set 𝑆. We define a homomorphism 𝜓1 :𝐻 →𝑆! so that for any ℎ,ℎ1 ∈𝐻

Let 𝜎 ∈𝑆! be the permutation which swaps 𝐴 and 𝑢 and leaves everything else invariant, and let ˆ𝜎 be its induced inner automorphism. Then 𝜓2 =ˆ𝜎𝜓1 is also a homomorphism. Now, for any 𝑎 ∈𝐴 it follows 𝜓1(𝑎) leaves both 𝐴 and 𝑢 fixed (the latter is by construction), and hence 𝜎 commutes with 𝜓1 and thus 𝜓2(𝑎) =ˆ𝜎𝜓1(𝑎) =𝜓1(𝑎). Hence 𝜓1𝜙 =𝜓2𝜙, and since 𝜙 is epic 𝜓1 =𝜓2. But thence follows that 𝜎 commutes with 𝜓1(ℎ) for all ℎ ∈𝐻, wherefore 𝜓1(ℎ) must leave 𝐴 fixed, and thus 𝜓1(ℎ) =ℎ𝐴 =𝐴 and thus ℎ ∈𝐴 for all ℎ ∈𝐻.