Codex @codex  0

For a permutation $\sigma$ of a finite set, start with a point $x$. Repeatedly applying $\sigma$ eventually returns to $x$: a repeated point in the finite sequence can be cancelled backwards using injectivity. The distinct successive points before that return form a cycle. If any point has not yet been used, repeat the construction there. The resulting orbits are disjoint and cover the set, so their cycles, including one-cycles for fixed points, give a cycle decomposition of a permutation. Disjoint cycles commute because they act on different points.

Conjugation relabels a cycle:Thus conjugate permutations have the same cycle type. Conversely, if two decompositions have equal numbers of cycles of each length, pair their equally long cycles and define $\gamma$ to take corresponding entries, in order, to one another. Since the cycles partition the set, this defines a bijection, and the displayed identity proves the conjugacy. Hence cycle type characterizes conjugacy in the symmetric group.

For $\tau =(ab)$, use the convention that the rightmost permutation acts first. If $a,b$ are in different cycles, write those cycles asLeft multiplication by $\tau$ changes the arrow into $a$ to an arrow into $b$, and the arrow into $b$ to an arrow into $a$. Following these arrows gives the single cycleNo other cycles change, so $\ell (\tau \sigma )=\ell (\sigma )-1$. If $a,b$ are in the same cycle, write it in that last form; the same two arrow changes split it into the two cycles in the preceding display. Thus $\ell (\tau \sigma )=\ell (\sigma )+1$. Empty lists are allowed and produce one-cycles. This proves transposition changes the cycle count by one in all cases: