Prove that if the identity is written as the product of $r$ transpositions, then $r$ is an even number Theorem. If the identity is written as the product of $r$ transpositions, $id=τ_1τ_2\dots τ_r$, then $r$ is an even number.
Proof. We will employ induction on $r$. A transposition cannot be the identity; hence, $r > 1$. If $r = 2$, then we are done. Suppose that $r > 2$. In this case the product of the last two transpositions, $τ_{r−1}τ_r$, must be one of the following cases:
$(ab)(ab) = id\\ (bc)(ab) = (ac)(bc)\\ (cd)(ab) = (ab)(cd)\\ (ac)(ab) = (ab)(bc),$
where $a$, $b$, $c$, and $d$ are distinct.
The first equation simply says that a transposition is its own inverse. If this case occurs, delete $τ_{r−1} τ_r$ from the product to obtain $id=τ_1τ_2···τ_{r−3}τ_{r−2}$. By induction $r − 2$ is even; hence, $r$ must be even.
In each of the other three cases, we can replace $τ_{r−1} τ_r$ with the right-hand side of the corresponding equation to obtain a new product of $r$ transpositions for the identity. In this new product the last occurrence of $a$ will be in the next-to-the-last transposition. We can continue this process with $τ_{r−2} τ_{r−1}$ to obtain either a product of $r − 2$ transpositions or a new product of $r$ transpositions where the last occurrence of $a$ is in $τ_{r−2}$. If the identity is the product of $r − 2$ transpositions, then again we are done, by our induction hypothesis; otherwise, we will repeat the procedure with $τ_{r−3}τ_{r−2}$.
At some point either we will have two adjacent, identical transpositions canceling each other out or $a$ will be shuffled so that it will appear only in the first transposition. However, the latter case cannot occur, because the identity would not fix $a$ in this instance. Therefore, the identity permutation must be the product of $r − 2$ transpositions and, again by our induction hypothesis, we are done.
Question: That is the proof written in the book, but unfortunately after more than $10$ times reading I don't understand the last two paragraph (i.e., from "In each of the other three cases"). Is there another simpler proof or could someone please clarify what is going on?
Thank you.
 A: You can identify a permutation with a permutation matrix, i.e., the one you get by permuting the columns of the identity matrix according to the permutation. Composition of permutations corresponds to matrix multiplication, and since the determinant of a transposition is $-1$, the determinant of a composition of $r$ transpositions is $(-1)^r$. So if it is the identity, then $(-1)^r=1$, implying that $r$ is even.
A: You could try using the signature of the permutations. If $id=\prod_{i=1}^r \tau_i$ where $\tau_i$ are transpositions, you have that $\varepsilon(id)=1$ and $\varepsilon(\tau_i)=(-1)$, therefore $(-1)^r=1\Rightarrow r$ even. 
I do not know if you studied signatures.
A: Notice that in each of the three given identities 
$(bc)(ab)=(ac)(bc)$
$(cd)(ab)=(ab)(cd)$
$(ac)(ab)=(ab)(bc)$
if you replace the left side with the right side, there is no longer an $a$ in the second parentheses.  This means that each time you apply any of these identities, the rightmost occurrence of $a$ is one transposition further to the left.
Eventually either you will get two identical adjacent transpositions $(ax)(ax)=id$, so they disappear and you can apply induction to the shorter string of transpositions; or you end up with $a$ only appearing in the leftmost transposition of the entire product of transpositions, but this is impossible since then the permutation doesn't fix $a$, and so is not the identity.
A: suppose the set of $n+1$ items on which the permutations act are mapped to the vertices of a regular simplex in n-dimensional Euclidean space, with its centroid at the origin of co-ordinates. 
for any pair of vertices $A$ and $B$ there is a unique hyperplane $\Gamma_{AB}$ which contains the remaining $n-1$ vertices and the midpoint of $AB$. 
reflection in $\Gamma_{AB}$ interchanges $A$ and $B$ leaving all the other points invariant. 
any such reflection reverses the orientation of the simplex. since the identity leaves orientation unchanged, there must be an even number of reflections in any sequence which simplifies to the identity transformation.
A: There are 4 possible forms of the last two transpositions in the equation. Replace each with their equivalent.
case 1: (ab)(ab)=ID
case 2: (bc)(ab)=(ac)(bc)
case 3: (cd)(ab)=(ab)(cd)
case 4: (ac)(ab)=(ab)(bc)
If case 1, we are done by induction hypothesis.
If cases 2-4, the last two transpositions are not equal ID. When we repeat this process with the third-to-last transposition and the "new" second-to-last transposition and obtain a case 2-4, we can now say that the three last transpositions are not equal to ID. Continue this process and obtain (a ?) as the first transposition, then we have a contradiction because every transposition to the right of our "new" (a ?) is not equal to the ID (but our original assumption was that the product of the transpositions was equal to the identity).
If case 1 happens at any point, then the permutation is even by the induction hypothesis. If case 1 never happens, then the there is a contradiction, so the permutation is not equal to ID. --> Any product of transpositions equal to ID must be even.
