Every odd permutation in $S_n$ can be written as a product of $2n+3$ transpositions?

My question is

If Set of permutation set $S_n$ for $n>3$

Prove Every odd permutation in Sn can be written as a product of 2n+3 transpositions and every even permutation as a product of $2n+8$ transpositions.

I am little bit confused.

If we consider $S_6$, there is a cycle $(1 4 5 6)$ which can be decomposed into $(1 6)(1 5)(1 4)$

Product of $3$ transpositions. this is odd permutation in $S_6$, how can it be written as product of $2n+3$ transpositions?

Did I misunderstand the question?

• So $(1 4 5 6) = (1 6)(1 5)(1 4) (1 6)(1 5)(1 4) (1 6)(1 5)(1 4) (1 6)(1 5)(1 4) (1 6)(1 5)(1 4)$, which is $2 \times 6 + 3$ transpositions. – MarkG Feb 15 '15 at 4:20
• If you can write a permutation as a product of $k$ transpositions, then you can also write it as a product of $k+2l$ transpositions for any $l \ge 0$, just by adjoining pairs $(1,2)(1,2)$ (at least when $n \ge 2$). – Derek Holt Feb 15 '15 at 8:25
• There's more than one way to decompose a permutation. – Akiva Weinberger Apr 18 '16 at 10:54