Question

Compute each product:

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

Short answer

The product is$\left(14532\right)$.

Step-by-step solution

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

We have the definition of cycle that,

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,

$\left(12\right)\left(53214\right)\left(23\right)$

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. Thus,we will start mapping from 1 in the second cycle from the right-hand.

$\left(12\right)\left(53214\right)\left(23\right)=\left(\begin{array}{ccccc}1& 2& 3& 4& 5\\ 4& 1& 2& 5& 3\end{array}\right)$

Here, 1 is mapped to 4.

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

Next,3 is mapped to 2.

Next, 4 is mapped to 5 and 5 is mapped back to 3.

To obtain final product

From step 1, we get,

$\left(\begin{array}{ccccc}1& 2& 3& 4& 5\\ 4& 1& 2& 5& 3\end{array}\right)$

Here, 1 is mapped to 4, 4, which is mapped to 5, 5, mapped to 3, 3, which is mapped to 2 and 2 is mapped back to 1.

The final product is expressed in the cycle notation as,

$\begin{array}{c}\left(12\right)\left(53214\right)\left(23\right)=\left(\begin{array}{ccccc}1& 2& 3& 4& 5\\ 4& 1& 2& 5& 3\end{array}\right)\\ =\left(14532\right)\end{array}$

Hence, the product is$\left(14532\right)$.