TITLE: Alternative proof that the parity of permutation is well defined?
QUESTION [36 upvotes]: I learned the following theorem about the properties of permutation from Gallian's Contemporary Abstract Algebra.

Theorem 5.4 $\;$ Always Even or Always Odd
If a permutation $\alpha$ can be expressed as a product of an even number of $2$-cycles, then every decomposition of $\alpha$ into a product of $2$-cycles must have an even number of $2$-cycles. In symbols, if
  $$
  \alpha = \beta_1 \beta_2 \dotsm \beta_r
  \quad\text{and}\quad
  \alpha = \gamma_1 \gamma_2 \dotsm \gamma_s
$$
  where the $\beta$'s and the $\gamma$'s are $2$-cycles, then $r$ and $s$ are both even or both odd.

When I tried to reconstruct the proof myself, I found that it suffices to prove the following lemma:

If $\epsilon=\beta_1\beta_2\cdots\beta_r$ where $\beta$'s are $2$-cycles, then $r$ is even. 

The original proof for this lemma uses the following key property of the product of $\beta_1\beta_2$:

The product can always be expressed in one of the following forms on the left:
  $$\begin{align}(ab)(ab)&=\epsilon\\
(ab)(ac)&=(bc)(ab)\\
(ab)(cd)&=(cd)(ab)\\
(ab)(bc)&=(bc)(ac)\end{align}$$

The proof for the lemma is based on such property and mathematical induction. I found it really hard to remind of such property, so I set it as an exercise to give another proof. However, I have no idea how to actually do it. 
So here is my question:

Does anybody know an alternative proof of the lemma or the theorem?

REPLY [2 votes]: There is another way, I believe it is more elementary.
Let $f\in\mathfrak{S}_n$ and $r \leq n$ the number of orbits of $f$, i.e. the number of cycles in the decomposition of $f$ in cycles with disjoint supports (the decomposition is unique up to order).
Let $\tau$ be a transposition. Show the following lemma: 
Lemma: the number of orbits of $\tau \circ f$ is either $r+1$ or $r-1$.
Hence, applying a transposition to $f$ can only increase or decrease the number of orbits by $1$.
Now consider a decomposition of $f$ in transpositions:
\begin{align*}
\tau_1 \circ \tau_2 \circ \dots \tau_p \circ \text{Id} &= f
\end{align*}
Let's write $p = k_1 + k_2$ with $k_1$ the number of transpositions which increased the number of orbits by one and $k_2$ the number of transpositions which decreased the number of orbits by one. 
Since we are going from $\text{Id}$, that has $n$ orbits, to $f$ that has $r$ orbits, the lemma implies that $k_2 \geq n-r$. Let's write $k_3 = k_2 - (n-r) \geq 0$, we have:
\begin{align*}
r &= n + k_1 - k_2 = n + k_1 - (n-r) - k_3 = r + k_1 - k_3
\end{align*}
Hence $k_1 = k_3$ and we have: 
\begin{align*}
p &= k_1 + k_2 \\
  &= k_1 + k_3 + (n-r)\\ 
  &= 2k_1 + (n-r)
\end{align*}
Hence:  $p \text{ mod } 2 = n-r \text{ mod } 2$. 
The parity of $p$ depends only on information present in $f$: that is, $r$, its number of orbits. 
Hence, any decomposition of $f$ into transpositions will have the same parity of number of transpositions: the parity of $n-r$.
Note that this proof was found from the fact that the optimal number of transpositions to decompose $f$ is $n-r$.