TITLE: Elements of $S_n$ which can not be product of $\leq n-2$ transpositions
QUESTION [9 upvotes]: It is well known that every element of $S_n$ can be written as a product of at most $n-1$ transpositions. This theorem is proved in all the books which discuss the permutation groups. But, I find that the following natural question is usually neither discussed nor put in Exercise Section in the book. 

Question: Does there always exists an element in $S_n$ which can not be product of $\leq n-2$ transpositions? Can we list the cycle-type of such elements?

The natural guess would be the $n$-cycles in $S_n$ would be the elements which answer the question. But I couldn't prove it. Can you help me? A hint would also be sufficient.

REPLY [3 votes]: It can be shown that: An element in $S_n$ can be written as a product of $\leq n-2$ transpositions if and only if the element is not an $n$-cycle.  So the answer to your question is: yes, there always exists elements in $S_n$ that cannot be written as a product of $\le n-2$ transpositions, and these elements are precisely the $n$-cycles.
To prove the above statement, you can examine how the number of cycles in a permutation changes when we compose the permutation with a transposition. If $g=(123)(45)(6) \in S_6$, then the number of cycles in $g$ is $c(g)=3$. You can work out examples to see that composing $g$ with any transposition $(i,j)$ reduces $c$ by 1 or increases $c$ by 1, according as whether $i$ and $j$ are in different or same cycles of $g$. For example, composing $g$ with $(14)$ will merge the two cycles in $g$ that contain 1 and 4, thereby reducing $c$ by 1, while composing $g$ with $(12)$ will split the cycle $(123)$ in $g$, thereby increasing $c$ by 1.  
Now, $g \in S_n$ is a product $g=t_1 \cdots t_k$ of transpositions iff $g t_k \cdots t_1$ is the identity (recall that a transposition is its own inverse).  By the argument in the previous paragraph, composing $g$ with $k$ transpositions can reduce or increase $c$ by at most $k$.  The identity permutation has $c=n$. We need to compose an $n$-cycle with transpositions until we get the identity permutation.  Starting with an $n$ cycle, which has $c=1$, the number of transpositions needed to split the $n$ cycle into $n$ 1-cycles (and thereby increase $c$ to $n$) is at least $n-1$ (by the argument in the previous paragraph).  Hence, an $n$-cycle cannot be expressed as a product of fewer than $n-1$ transpositions.