Hi I have the following questions:
Calculate the following compositions of permutations on $A=\{0,1,2,3\}$
- $(12)(102)$
Ans:$(02)$
- $(01)(23)(0123)$
Ans:$(02)$
- $(123)(32)$
Ans:$(13)$
Let $A={1,2,3,4,5}$ and let $F$ and $G$ be permutations on $A$ defined as follows:
$$F=\{(1,1),(2,5),(3,3),(4,4),(5,2)\}$$
$$G=\{(1,3),(2,2),(3,1),(4,5),(5,4)\}$$
Write your answers in cycle notation for the part (i) and (ii)
(i) write down F and G
Ans: $F=(25)$ and $G=(13)(45)$
(ii) Write down the composition of $GF$
Ans: $GF = (13)(254)$
Can you help advise how the answer was derived?
I will help you with the first problem. I will do two using an algorithmic approach, leaving the last one for you to figure out.
Let's look at the composition $(12)(102)$. It seems that your convention is to compose left to right, so we start on the left. (Warning: as far as I know, the standard is to compose right to left, so be wary when seeking general resources.) $0$ goes to $0$ in the first permutation, and $0$ goes to $2$ in the second. So in the composition, $0$ goes to $2$. $1$ goes to $2$ in the first permutation, and $2$ goes to $1$ in the second. So that explains why $1$ is fixed. $2$ goes to $1$ in the first permutation, and $1$ goes to $0$, so $2$ goes to $0$ overall. This leaves $3$ fixed (which should be obvious from the outset). Thus, the composition is $(02)$.
Next let's look at $(01)(23)(0123)$. Walking through the permutations, $0$ goes to $1$, then $1$ goes to $1$, and then $1$ goes to $2$. So in the composition, $0$ goes to $2$. $1$ goes to $0$, then $0$ goes to $0$, then $0$ goes to $1$. So $1$ is fixed. $2$ goes to $2$, then $2$ goes to $3$, then $3$ goes to $0$. So in the composition, $2$ goes to $0$. There is only one place $3$ can go: $3$. So the composition is $(02)$.
(Edited September 10, 2014)
Since you corrected the typo, let's take a look at the second problem.
The notation for $F$ reflects the fact that $2$ and $5$ are the only swapped variables. The notation for $G$ reflects the fact that $1$ and $3$ form a two period cycle and that $4$ and $5$ form a two period cycle. Since $2$ is fixed, it does not appear.
In composing $F$ and $G$, we use the same technique as before. Going left to right, $1$ goes to $3$, and then $3$ stays fixed. So in the composition, $1$ goes to $3$. $2$ goes to $2$, then $2$ goes to $5$. So in the composition, $2$ goes to $5$. Likewise, we see that $3$ goes to $1$ and then is fixed at $1$, so in the composition $3$ goes to $1$. $4$ goes to $5$, and then $5$ goes to $2$, so in the composition $4$ goes to $2$. Last we have $5$, which must go to the only open symbol: $4$.
It is clear that $1$ and $3$ form a two-cycle and that $2, 5,$ and $4$ form a three way cycle, in that order.