Why is $(12345) = (54)(53)(52)(51)$? I'm trying to understand the concept of permutations being decomposed into transpositions in group theory. In Pinter's Book of Abstract Algebra pg. 83, he says
$(12345) = (54)(53)(52)(51)$
I just can't make any sense of this. The permutation $(12345)$ is the identity, right? So if I apply those transpositions to $(12345)$ starting from the right, I should end up with $(12345)$ again, correct?
When I write out each permutation in full, I get
$(52341)\\
(15342)\\
(12543)\\
(12354)$
But when I trace what happens to each of the numbers 1, 2, 3, 4 and 5, I end up with $(23451)$. For instance, 1 goes to 5, then 2, then 2, then 2. And 2 goes to 2, then 5, then 3, then 3. And so on.
He provides a number of other decompositions, and I get the same end result from all of them.
What am I doing wrong?
 A: This is cycle notation, which is ideal for many types of calculations with permutations. For example, the notation $(1\,2\,3\,4\,5)$ stands for the transformation
that sends
$$
1 \mapsto 2 \mapsto 3 \mapsto 4 \mapsto 5 \mapsto 1.
$$
It's not unique. In fact, for a $5$-cycle such as this, there are $5$ ways to represent the permutation, depending on which one you put first. Also, any symbol that doesn't appear in a cycle is understood to be fixed by that cycle.
Now, permutations are bijective functions, so you compose them to form the permutation group. This is denoted by concatenation. For example, just composing the first two (reading from right to left, as you do with functions), let's calculate:
$$
(5\,2)(5\,1).
$$
Starting on the right, the first permutation maps $5 \mapsto 1$, then the second permutation leaves $1$ fixed. So in the composition, $5 \mapsto 1$. Repeating this for other elements, you can see that
\begin{align*}
&1 \mapsto 5 \mapsto 2 \\
&2 \mapsto 2 \mapsto 5 \\
&5 \mapsto 1 \mapsto 1 
\end{align*}
Any other symbol is fixed (e.g. $3 \mapsto 3 \mapsto 3$ under the composition).
Thus,
$$
(5\,2)(5\,1) = (5\,1\,2)
$$
as cycles. With this understanding you should be able to reproduce the other calculations.
A: No, the permutation (1 2 3 4 5) is not the identity.  It is the cyclic permutation that sends: 1 to 2, 2 to 3, 3 to 4, 4 to 5, 5 to 1.
To simplify the explanation of decomposing into transpositions, let's instead use a simpler cyclic permutation: (1 2 3).  It sends 1 to 2, 2 to 3, and 3 to 1.
Now consider the product of transpositions: (1 3) (2 3).  It sends:
*) 1 to 3 to 2; i.e., ultimately, 1 to 2;
*) 3 to 1;
*) 2 to 3.
I.e., it has the same ultimate effect as permutation (1  2  3).
You will get a clear exposition to groups and, in particular, to permutations, from Part 1 of Abel's theorem in problems and solutions.
