Chapter 14, Problem 1E

(a)

To determine

The value of the permutations $f\circ g$ and $g\circ f$.

Answer to Problem 1E

The permutations $\underset{\_}{f\circ g=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 2& 5& 3& 4& 1& 6\end{array}\right)}$ and $\underset{\_}{g\circ f=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 1& 2& 5& 3& 4& 6\end{array}\right)}$.

Explanation of Solution

The given permutations are, $f=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 6& 4& 2& 1& 5& 3\end{array}\right)$ and $g=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 3& 5& 6& 2& 4& 1\end{array}\right)$.

Here, $\left(f\circ g\right)\left(1\right)=2$, $\left(f\circ g\right)\left(2\right)=5$, $\left(f\circ g\right)\left(3\right)=3$, $\left(f\circ g\right)\left(4\right)=4$, $\left(f\circ g\right)\left(5\right)=1$ and $\left(f\circ g\right)\left(6\right)=6$.

Thus, $f\circ g=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 2& 5& 3& 4& 1& 6\end{array}\right)$.

Similarly, $\left(g\circ f\right)\left(1\right)=1$, $\left(g\circ f\right)\left(2\right)=2$, $\left(g\circ f\right)\left(3\right)=5$, $\left(g\circ f\right)\left(4\right)=3$, $\left(g\circ f\right)\left(5\right)=4$ and $\left(g\circ f\right)\left(6\right)=6$.

Thus, $g\circ f=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 1& 2& 5& 3& 4& 6\end{array}\right)$.

Therefore, the permutations are $\underset{\_}{f\circ g=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 2& 5& 3& 4& 1& 6\end{array}\right)}$ and $\underset{\_}{g\circ f=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 1& 2& 5& 3& 4& 6\end{array}\right)}$.

(b)

To determine

The value of the permutation $f^{-1}$ and $g^{-1}$.

Answer to Problem 1E

The permutations $\underset{\_}{f^{-1}=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 4& 3& 6& 2& 5& 1\end{array}\right)}$ and $\underset{\_}{g^{-1}=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 6& 4& 1& 5& 2& 3\end{array}\right)}$.

Explanation of Solution

The given permutations are, $f=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 6& 4& 2& 1& 5& 3\end{array}\right)$ and $g=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 3& 5& 6& 2& 4& 1\end{array}\right)$.

Obtain the permutation $f^{-1}$ from f as follows:

Interchange the rows of the permutation f.

Then, the permutation will be $\left(\begin{array}{cccccc}6& 4& 2& 1& 5& 3\\ 1& 2& 3& 4& 5& 6\end{array}\right)$.

Arrange the columns of the permutation $f^{-1}$ in ascending order.

Thus, $f^{-1}=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 4& 3& 6& 2& 5& 1\end{array}\right)$.

Obtain the permutation $g^{-1}$ from g as follows:

Interchange the rows of the permutation g.

Then, the permutation will be $\left(\begin{array}{cccccc}3& 5& 6& 2& 4& 1\\ 1& 2& 3& 4& 5& 6\end{array}\right)$.

Arrange the columns of the permutation $g^{-1}$ in ascending order.

Thus, $g^{-1}=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 6& 4& 1& 5& 2& 3\end{array}\right)$.

Therefore, the permutations are $\underset{\_}{f^{-1}=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 4& 3& 6& 2& 5& 1\end{array}\right)}$ and $\underset{\_}{g^{-1}=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 6& 4& 1& 5& 2& 3\end{array}\right)}$.

(c)

To determine

The values of the permutation $f^2$ and $f^5$.

Answer to Problem 1E

The permutations are $\underset{\_}{f^2=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 3& 1& 4& 6& 5& 2\end{array}\right)}$ and $\underset{\_}{f^5=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 1& 2& 3& 4& 5& 6\end{array}\right)}$.

Explanation of Solution

The given permutation is, $f=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 6& 4& 2& 1& 5& 3\end{array}\right)$.

Obtain the permutation $f^2$ from f as follows:

Here, $f^2$ means $f\circ f$.

Clearly, $\left(f\circ f\right)\left(1\right)=3$, $\left(f\circ f\right)\left(2\right)=1$, $\left(f\circ f\right)\left(3\right)=4$, $\left(f\circ f\right)\left(4\right)=1$, $\left(f\circ f\right)\left(5\right)=5$ and $\left(f\circ f\right)\left(6\right)=3$.

Therefore, $f\circ f=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 3& 1& 4& 6& 5& 2\end{array}\right)$.

That is, $f^2=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 3& 1& 4& 6& 5& 2\end{array}\right)$.

Obtain the permutation $f^5$ as follows:

Here, $f^5$ means $f\circ f^2\circ f^2$.

Clearly, $\left(f^2\circ f^2\right)\left(1\right)=4$, $\left(f^2\circ f^2\right)\left(2\right)=3$, $\left(f\circ f\right)\left(3\right)=6$, $\left(f\circ f\right)\left(4\right)=2$, $\left(f\circ f\right)\left(5\right)=5$ and $\left(f\circ f\right)\left(6\right)=1$.

Thus, $f^2\circ f^2=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 4& 3& 6& 2& 5& 1\end{array}\right)$.

Now, compute $f\circ f^2\circ f^2$.

Clearly, $\left(f\circ f^2\circ f^2\right)\left(1\right)=1$, $\left(f\circ f^2\circ f^2\right)\left(2\right)=2$, $\left(f\circ f^2\circ f^2\right)\left(3\right)=3$, $\left(f\circ f^2\circ f^2\right)\left(4\right)=4$, $\left(f\circ f^2\circ f^2\right)\left(5\right)=5$ and $\left(f\circ f^2\circ f^2\right)\left(6\right)=6$.

Thus, $f\circ f^2\circ f^2=f^5=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 1& 2& 3& 4& 5& 6\end{array}\right)$.

Therefore, the permutations are $\underset{\_}{f^2=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 3& 1& 4& 6& 5& 2\end{array}\right)}$ and $\underset{\_}{f^5=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 1& 2& 3& 4& 5& 6\end{array}\right)}$.

(d)

To determine

The value of the permutation $f\circ g\circ f$.

Answer to Problem 1E

The permutations is $\underset{\_}{f\circ g\circ f=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 6& 4& 5& 2& 1& 3\end{array}\right)}$.

Explanation of Solution

The given permutations are, $f=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 6& 4& 2& 1& 5& 3\end{array}\right)$ and $g=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 3& 5& 6& 2& 4& 1\end{array}\right)$.

Obtain the permutation $f\circ g\circ f$ as follows:

From part (a), $g\circ f=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 2& 5& 3& 4& 1& 6\end{array}\right)$.

Clearly, $\left(f\circ g\circ f\right)\left(1\right)=6$, $\left(f\circ g\circ f\right)\left(2\right)=4$, $\left(f\circ g\circ f\right)\left(3\right)=5$, $\left(f\circ g\circ f\right)\left(4\right)=2$, $\left(f\circ g\circ f\right)\left(5\right)=1$ and $\left(f\circ g\circ f\right)\left(6\right)=3$.

Thus, $f\circ g\circ f=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 6& 4& 5& 2& 1& 3\end{array}\right)$.

Therefore, the permutations is $\underset{\_}{f\circ g\circ f=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 6& 4& 5& 2& 1& 3\end{array}\right)}$.

(e)

To determine

The value of the permutation $g^3$ and $f\circ g^3\circ f^{-1}$.

Answer to Problem 1E

The permutations are $\underset{\_}{g^3=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 1& 2& 3& 4& 5& 6\end{array}\right)}$ and $\underset{\_}{f\circ g^3\circ f^{-1}=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 1& 2& 3& 4& 5& 6\end{array}\right)}$.

Explanation of Solution

The given permutations are, $f=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 6& 4& 2& 1& 5& 3\end{array}\right)$ and $g=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 3& 5& 6& 2& 4& 1\end{array}\right)$.

Obtain the permutation $g^3$ as follows:

Here, $g^3$ means $g\circ g\circ g$.

Clearly, $\left(g\circ g\right)\left(1\right)=6$, $\left(g\circ g\right)\left(2\right)=4$, $\left(g\circ g\right)\left(3\right)=1$, $\left(g\circ g\right)\left(4\right)=5$, $\left(g\circ g\right)\left(5\right)=2$ and $\left(g\circ g\right)\left(6\right)=3$.

Thus, $g\circ g=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 6& 4& 1& 5& 2& 3\end{array}\right)$.

Similarly, $\left(g\circ g\circ g\right)\left(1\right)=1$, $\left(g\circ g\circ g\right)\left(2\right)=2$, $\left(g\circ g\circ g\right)\left(3\right)=3$, $\left(g\circ g\circ g\right)\left(4\right)=4$, $\left(g\circ g\circ g\right)\left(5\right)=5$ and $\left(g\circ g\circ g\right)\left(6\right)=6$.

Thus, $g\circ g\circ g=g^3=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 1& 2& 3& 4& 5& 6\end{array}\right)$.

From part (b), the permutation $f^{-1}=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 4& 3& 6& 2& 5& 1\end{array}\right)$.

Now, obtain the permutation $f\circ g^3\circ f^{-1}$ as follows:

Clearly, $\left(g^3\circ f^{-1}\right)\left(1\right)=6$, $\left(g^3\circ f^{-1}\right)\left(2\right)=4$, $\left(g^3\circ f^{-1}\right)\left(3\right)=1$, $\left(g^3\circ f^{-1}\right)\left(4\right)=5$, $\left(g^3\circ f^{-1}\right)\left(5\right)=2$ and $\left(g^3\circ f^{-1}\right)\left(6\right)=3$.

Thus, $g^3\circ f^{-1}=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 4& 3& 6& 2& 5& 1\end{array}\right)$.

Similarly, $\left(f\circ g^3\circ f^{-1}\right)\left(1\right)=1$, $\left(f\circ g^3\circ f^{-1}\right)\left(2\right)=2$, $\left(f\circ g^3\circ f^{-1}\right)\left(3\right)=3$, $\left(f\circ g^3\circ f^{-1}\right)\left(4\right)=4$, $\left(f\circ g^3\circ f^{-1}\right)\left(5\right)=5$ and $\left(f\circ g^3\circ f^{-1}\right)\left(6\right)=6$.

Thus, $f\circ g^3\circ f^{-1}=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 1& 2& 3& 4& 5& 6\end{array}\right)$.

Therefore, the permutations are $\underset{\_}{g^3=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 1& 2& 3& 4& 5& 6\end{array}\right)}$ and $\underset{\_}{f\circ g^3\circ f^{-1}=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 1& 2& 3& 4& 5& 6\end{array}\right)}$.