Trouble understanding transpositions This problem comes from the transposition area of the Wikipedia page for symmetric group (link).
We have $g=(1 2 5)(3 4)$. It then says that g can also be written as $(1 2)(2 5)(3 4)$, but I don't understand why. I follow why g sends 1 to 2, 2 to 5, 3 to 4, 4 to 3, and 5 to 1 under its original representation. But under the second representation, wouldn't 5 go to 2? I can slightly follow that 5 going to 2 means it goes to 1, but then why doesn't 2 to go to 1, but instead it goes to 5?
 A: Let's see what the composition $(12)(25)(34)$ does to each number:
$$
1 \overset{(34)}{\to} 1 \overset{(25)}{\to} 1 \overset{(12)}{\to}2\\
2 \overset{(34)}{\to} 2 \overset{(25)}{\to} 5 \overset{(12)}{\to}5\\
3 \overset{(34)}{\to} 4 \overset{(25)}{\to} 4 \overset{(12)}{\to}4\\
4 \overset{(34)}{\to} 3 \overset{(25)}{\to} 3 \overset{(12)}{\to}3\\
5 \overset{(34)}{\to} 5 \overset{(25)}{\to} 2 \overset{(12)}{\to}1
$$
Note that, just as in the composition of functions, our transpositions are applied from right to left.
A: $(12)(25)(34)$ is read right to left. So $3$ goes to $4$, $4$ goes to $3$, $2$ goes to $5$, $5$ goes to $2$ which then goes to $1$ and $1$ goes to $2$.
A: Since notations can change across books, perhaps you may want to see how transposition product is introduced on Wiki.

Transpositions
$$(a~b~c~d~\ldots ~y~z)=(a~b)\cdot (b~c~d~\ldots ~y~z)$$
  This means the initial request is to move $a$ to $b$, $b$ to $c$, $y$ to $z$ and finally $z$ to $a$. Instead one may roll the elements keeping $a$ where it is by executing the right factor first.

In this case, $(3~4)$ is independent from $(1~2~5)=(1~2)(2~5)$.
Mneumonic skill: you may find it easy to remember that on the RHS, we are just writing the same element twice across the dot sign, and keep the order unchanged.
