We compute hgh^-1(x) by distinguishing two cases.

• x = h(a_i) for some 1 i m:
  hgh^-1(h(a_i)) = hg(a_i) = h(a_{i + 1})

  (with the convention that a_{m + 1} = a₁).

• If x is not in {h(a₁ ), ..., h(a _m)} then h^-1(x) is not in {a₁, ..., a_m}, so that g(h^-1(x)) = h^-1(x) and consequently hgh^-1(x) = x.

We conclude that hgh^-1 = (h(a₁ ), ..., h(a_m)).

The second item of the lemma follows once you realize that in the product

the pairs h^-1h cancel, so that hg₁g₂ ··· g_th^-1 is what remains.

In particular, for cycles c_i, the first item of the lemma then shows that the product hc₁h^-1 hc₂h^-1 ··· hc_th^-1 is the product of the cycles hc_ih^-1.

If cycles have disjoint supports, then their conjugates also have disjoint supports: The support of hc_ih^-1 is h(supp(c_i)) (see the first item of the lemma), so that the supports of hc₁h^-1, hc₂h^-1, ..., hc_th^-1 are the sets h(supp(c₁)), h(supp(c₂)), ..., h(supp(c_t)). Since h is a bijection, these sets are disjoint if the sets supp(c₁), ..., supp(c_t) are disjoint.