Obviously, there are lots of answers already, but I thought I'd give the proof of (5) as given by Jordan already in 1870 in his Traité des substitutions -- this has the benefit of being quite clear to read, and cuts directly into the "conceptual".

The following is my translation Theorem 77, its proof, and the paragraph above it (pp. 64--65), with some update of the words used (e.g. permutation instead of "substitution"):

A cycle on $p$ letters $a, b, c, \dots$ is clearly the product of the $p-1$ transpositions $(ab), (ac), \dots$. Hence an arbitrary permutation on $k$ letters and containing $n$ cycles is the product of $k-n$ transpositions.

Theorem: Let $S, T$ be two permutations which are respectively the product of $\alpha$ and $\beta$ transpositions. The number of successive transpositions in the product $ST$ is even or odd, according as to whether $\alpha + \beta$ is even or odd.

Proof: As any permutation is the product of transpositions, it suffices to show the theorem when $T$ is a transposition, in which case $\beta = 1$.

To make the idea clear, let $S = (abcde)(fg)$. If $T$ transposes two letters which appear in the same cycle, such as $a$ and $c$, then we have $$ ST = (ab)(cde)(fg), $$ which is a permutation containing one cycle more than $S$, and which, accordingly, is a product of $\alpha-1$ transpositions. If on the other hand $T$ transposes two letters $a, f$ appearing in different cycles, then $ST=(abcdefg)$ will contain one cycle less than $S$, and will hence be the product of $\alpha+1$ transpositions. In either case, the number of transpositions in the product $ST$ will be congruent to $\alpha + 1 \pmod 2$.

I think this counts more as the "traditional" proof than the determinant-proof, as it appears in the first book on group theory ever published :-) Of course, it does fall into "Using a decomposition into disjoint cycles, one can simply compute what happens when multiplying by a transposition", but it doesn't strike me as magic (and I don't think it struck Jordan as magic either!)

Obviously, there are lots of answers already, but I thought I'd give the proof of (5) as given by Jordan already in 1870 in his Traité des substitutions -- this has the benefit of being quite clear to read, and cuts directly into the "conceptual".

The following is my translation of his Theorem 77, its proof, and the paragraph above it (pp. 64--65), with some update of the words used (e.g. permutation instead of "substitution"):

A cycle on $p$ letters $a, b, c, \dots$ is clearly the product of the $p-1$ transpositions $(ab), (ac), \dots$. Hence an arbitrary permutation on $k$ letters and containing $n$ cycles is the product of $k-n$ transpositions.

Theorem: Let $S, T$ be two permutations which are respectively the product of $\alpha$ and $\beta$ transpositions. The number of successive transpositions in the product $ST$ is even or odd, according as to whether $\alpha + \beta$ is even or odd.

Proof: As any permutation is the product of transpositions, it suffices to show the theorem when $T$ is a transposition, in which case $\beta = 1$.

To make the idea clear, let $S = (abcde)(fg)$. If $T$ transposes two letters which appear in the same cycle, such as $a$ and $c$, then we have $$ ST = (ab)(cde)(fg), $$ which is a permutation containing one cycle more than $S$, and which, accordingly, is a product of $\alpha-1$ transpositions. If on the other hand $T$ transposes two letters $a, f$ appearing in different cycles, then $ST=(abcdefg)$ will contain one cycle less than $S$, and will hence be the product of $\alpha+1$ transpositions. In either case, the number of transpositions in the product $ST$ will be congruent to $\alpha + 1 \pmod 2$.

I think this counts more as the "traditional" proof than the determinant-proof, as it appears in the first book on group theory ever published :-) Of course, it does fall into "Using a decomposition into disjoint cycles, one can simply compute what happens when multiplying by a transposition", but it doesn't strike me as magic (and I don't think it struck Jordan as magic either!)

Obviously, there are lots of answers already, but I thought I'd give the proof of (5) as given by Jordan already in 1870 in his Traité des substitutions -- this has the benefit of being quite clear to read, and cuts directly into the "conceptual".

The following is my translation Theorem 77, its proof, and the paragraph above it (pp. 64--65), with some update of the words used (e.g. permutation instead of "substitution"):

A cycle on $p$ letters $a, b, c, \dots$ is clearly the product of the $p-1$ transpositions $(ab), (ac), \dots$. Hence an arbitrary permutation on $k$ letters and containing $n$ cycles is the product of $k-n$ transpositions.

Theorem: Let $S, T$ be two permutations which are respectively the product of $\alpha$ and $\beta$ transpositions. The number of successive transpositions in the product $ST$ is even or odd, according as to whether $\alpha + \beta$ is even or odd.

Proof: As any permutation is the product of transpositions, it suffices to show the theorem when $T$ is a transposition, in which case $\beta = 1$.

To make the idea clear, let $S = (abcde)(fg)$. If $T$ transposes two letters which appear in the same cycle, such as $a$ and $c$, then we have $$ ST = (ab)(cde)(fg), $$ which is a permutation containing one cycle more than $S$, and which, accordingly, is a product of $\alpha-1$ transpositions. If on the other hand $T$ transposes two letters $a, f$ appearing in different cycles, then $ST=(abcdefg)$ will contain one cycle less than $S$, and will hence be the product of $\alpha+1$ transpositions. In either case, the number of transpositions in the product $ST$ will be congruent to $\alpha + 1 \pmod 2$.

I think this counts more as the "traditional" proof than the determinant-proof, as it appears in the first book on group theory ever published :-)

Obviously, there are lots of answers already, but I thought I'd give the proof of (5) as given by Jordan already in 1870 in his Traité des substitutions -- this has the benefit of being quite clear to read, and cuts directly into the "conceptual".

The following is my translation Theorem 77, its proof, and the paragraph above it (pp. 64--65), with some update of the words used (e.g. permutation instead of "substitution"):

A cycle on $p$ letters $a, b, c, \dots$ is clearly the product of the $p-1$ transpositions $(ab), (ac), \dots$. Hence an arbitrary permutation on $k$ letters and containing $n$ cycles is the product of $k-n$ transpositions.

Theorem: Let $S, T$ be two permutations which are respectively the product of $\alpha$ and $\beta$ transpositions. The number of successive transpositions in the product $ST$ is even or odd, according as to whether $\alpha + \beta$ is even or odd.

Proof: As any permutation is the product of transpositions, it suffices to show the theorem when $T$ is a transposition, in which case $\beta = 1$.

To make the idea clear, let $S = (abcde)(fg)$. If $T$ transposes two letters which appear in the same cycle, such as $a$ and $c$, then we have $$ ST = (ab)(cde)(fg), $$ which is a permutation containing one cycle more than $S$, and which, accordingly, is a product of $\alpha-1$ transpositions. If on the other hand $T$ transposes two letters $a, f$ appearing in different cycles, then $ST=(abcdefg)$ will contain one cycle less than $S$, and will hence be the product of $\alpha+1$ transpositions. In either case, the number of transpositions in the product $ST$ will be congruent to $\alpha + 1 \pmod 2$.

I think this counts more as the "traditional" proof than the determinant-proof, as it appears in the first book on group theory ever published :-) Of course, it does fall into "Using a decomposition into disjoint cycles, one can simply compute what happens when multiplying by a transposition", but it doesn't strike me as magic (and I don't think it struck Jordan as magic either!)