How to multiply permutations I didn't understand the rules of multiplying  permutations.
I'll be glad if you can explain me...
For example: we have $f=(135)(27),g=(1254)(68)$. How do I calculate $f\cdot g$??
Thank you!
 A: I’m going to assume that you evaluate products from left to right; note that many people use the opposite convention. 
Here we have the product $(135)(27)(1254)(68)$. First see what it does to $1$: the cycle $(135)$ sends $1$ to $3$, the transposition $(27)$ does nothing to the $3$, the cycle $(1254)$ does nothing to the $3$, and the transposition $(68)$ does nothing to the $3$, so the net effect is to send $1$ to $3$.
Now let’s see what the product does to $3$. $(135)$ sends $3$ to $5$; $5$ is left unchanged by $(27)$, and $(1254)$ then sends $5$ to $4$, which is left unchanged by $(68)$. The net effect is to send $3$ to $4$.
Next we’ll see what the product does to $4$. Nothing changes it until we get to $(1254)$, which sends it to $1$, and $1$ is left alone by $(68)$, so the product sends $4$ to $1$. Notice that this closes a cycle, $1\to 3\to 4\to 1$.
Now start over with $2$, the smallest number whose image under $f\cdot g$ we’ve not yet found. $(135)$ does nothing to it, $(27)$ sends it to $7$, and $(1254)$ and $(68)$ do nothing to the $7$, so the net effect is to send $2$ to $7$. Next we see what the product does to $7$: $(135)$ does nothing, $(27)$ sends it to $2$, $(1254)$ sends that $2$ to $5$, and $(68)$ does nothing to the $5$, so $f\cdot g$ sends $7$ to $5$. $(135)$ sends $5$ to $1$, which is left alone by $(27)$ and then sent to $2$ by $(1254)$; $2$ is left alone by $(68)$, so $f\cdot g$ sends $5$ to $2$. This closes another cycle: $2\to 7\to 5\to 2$.
We now know what $f\cdot g$ does to $1,2,3,4,5$, and $7$, so we move on to $6$. $(135)$, $(27)$, and $(1254)$ do nothing to $6$, and $(68)$ sends it to $8$. $(135)$, $(27)$, and $(1254)$ also do nothing to $8$, which is sent to $6$ by $(68)$, and we’ve closed off one last cycle: $6\to 8\to 6$.
Putting the pieces together, we can write $f\cdot g$ as the product $(134)(275)(68)$.
Alternatively, you could work out $f\cdot g$ in two-line notation. First see where it sends $1$, then where it sends $2$, and so on up through $8$; you’ll be doing the same work that I did above, but in a different order, and you’ll find that $f\cdot g$ is
$$\binom{1\,2\,3\,4\,5\,6\,7\,8}{3\,7\,4\,1\,2\,8\,5\,6}$$
in two-line notation.
A: In the group of permutations, the multiplication means composition. So $f \cdot g := g \circ f$ (first do $f$ then do $g$). If your book or class has defined it in the opposite way but for the purposes of this response lets assume it is $f \cdot g := g \circ f$. The way I calculate this quickly is to make tables:
Starting Point (assuming the set has 8 elements): (1 2 3 4 5 6 7 8)
Apply $f$: (5 7 1 4 3 6 2 8)
(If you don't see how $f$ corresponds to this guy speak up.
Apply $g$ to previous line: (4 5 1 3 7 8 2 6)
List these three 8-tuples on top of one another and see if you can correspond $f$ to the first transition and $g$ to the second transition.
