Question

Compute each product:

$\begin{array}{l}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{\left(}\mathbf{12}\mathbf{\right)}\mathbf{\left(}\mathbf{23}\mathbf{\right)}\mathbf{\left(}\mathbf{34}\mathbf{\right)}\mathbf{}\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{\left(}\mathbf{246}\mathbf{\right)}\mathbf{\left(}\mathbf{147}\mathbf{\right)}\mathbf{\left(}\mathbf{135}\mathbf{\right)}\\ \\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{\left(}\mathbf{12}\mathbf{\right)}\mathbf{\left(}\mathbf{53214}\mathbf{\right)}\mathbf{\left(}\mathbf{23}\mathbf{\right)}\mathbf{\left(}\mathbf{d}\mathbf{\right)}\mathbf{\left(}\mathbf{1234}\mathbf{\right)}\mathbf{\left(}\mathbf{2345}\mathbf{\right)}\end{array}$

Short answer

The product is$\left(1234\right)$.

Step-by-step solution

Step by Step Solution Step 1: To compute the permutation from the given cycles

We have thedefinition of cyclethat,

Let $a_{1,}{a}_2,\dots ,a_k$be the distinct elements of the set $\{1,2,3,\dots ,n\}$.

Then$(a_1{a}_2\dots a_k)$, denotes the permutation in $S_n$that maps $a_1$to $a_2$, $a_2$to $a_3$,....,$a_{k-1}$to $a_k$and $a_k$to $a_1$and every other elements of set maps to themselves.

$(a_1{a}_2\dots a_k)$is called a cycle of length k or k cycle.

Now, the given cycle is:

$\left(12\right)\left(23\right)\left(34\right)$

We have to compute the product of these cycles to obtain the permutation.

Here, the given product is a composition of functions. So, the right-hand permutation is performed first.

$\left(12\right)\left(23\right)\left(34\right)=\left(\begin{array}{cccc}1& 2& 3& 4\\ 2& 3& 4& 1\end{array}\right)$

Here, 3 is mapped to 4.

Next, 4 is mapped to 3, 3, which is mapped to 2, and 2 is mapped to 1. So, 4 is mapped to 1.

Next, 2 is mapped to 3 and 1 is mapped to 2.

To obtain final product

From step 1, we get:

$\left(\begin{array}{cccc}1& 2& 3& 4\\ 2& 3& 4& 1\end{array}\right)$

Here, 1 is mapped to 2, 2, which is mapped to 3, 3, mapped to 4, and 4 is mapped back to 1.

The final product is expressed in the cycle notation as:

$$\begin{gathered} \left(12\right)\left(23\right)\left(34\right)=\left(\begin{array}{cccc}1& 2& 3& 4\\ 2& 3& 4& 1\end{array}\right) \\ =\left(1234\right) \end{gathered}$$

Hence, the product is$\left(1234\right)$.