Question

Compute each product:

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

Short answer

The product is $\left(12453\right)$.

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 or cycle.

Now, the given cycle is,

$\left(1234\right)\left(2345\right)$

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 right-hand.

$\left(1234\right)\left(2345\right)=\left(\begin{array}{ccccc}1& 2& 3& 4& 5\\ 2& 4& 1& 5& 3\end{array}\right)$

Here, 1 is mapped to 2.

Next, 2 is mapped to 3, 3 is mapped to 4. So, 2 is mapped to 4

Next, 3 is mapped to 4, 4 is mapped to 1. So, 3 is mapped to 1.

Next, 4 is mapped to 5.

Next, 5 is mapped to 2, 2, mapped to 3. So, 5 is mapped to 3.

To obtain final product

From step 1, we get,

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

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

The final product is expressed in the cycle notation as,

$\begin{array}{c}\left(1234\right)\left(2345\right)=\left(\begin{array}{ccccc}1& 2& 3& 4& 5\\ 2& 4& 1& 5& 3\end{array}\right)\\ =\left(12453\right)\end{array}$

Hence, the product is$\left(12453\right)$