# Thread: group and subgroup problem

1. ## group and subgroup problem

let $\rho , \sigma , \tau$ be the element of $S_{6}$ given by

$\rho=(26) , \sigma = (123) , \tau = (45)$

(1). Find $\rho \sigma \tau$and $\rho^{-1} \sigma^{-1} \tau^{-1}$
and verify $( \rho \sigma \tau )^{-1}= \rho^{-1} \sigma^{-1} \tau^{-1}$
(2). find the order of $\sigma \tau$and $\rho \sigma$

2. Originally Posted by ramiee2010
let $\rho , \sigma , \tau$ be the element of $S_{6}$ given by

$\rho=(26) , \sigma = (123) , \tau = (45)$

(1). Find $\rho \sigma \tau$and $\rho^{-1} \sigma^{-1} \tau^{-1}$
and verify $( \rho \sigma \tau )^{-1}= \rho^{-1} \sigma^{-1} \tau^{-1}$
(2). find the order of $\sigma \tau$and $\rho \sigma$
What is it that you do not understand about this question? It is a rather hard question to answer over the internet as it is about understanding, not about a simple `trick'. However, you should remember that what you are doing is the same as using a normal permutation. That is,

$(123) = \left(\begin{array}{cccccc} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 1 & 4 & 5 & 6 \end{array}\right)$.

To multiply to disjoint cycles, you look at where the preceeding permutation takes the element you are looking at.

So, for example,

$2(253) = 5, 5(253)=3, 3(253)=2$ and all the rest are kept as they are, and so $1(253)(15) = 1(15)=5, 2(253)(15) = 5(15) = 1$ etcetera.

Does that make sense?

Now, to find the inverse of a (single) disjoint cycle, you should notice that if $a_i \rightarrow a_{i+1}$ then you want $a_{i+1}\rightarrow a_i$ in the inverse. This is equivalent to keeping the first element fixed and flipping all the others round,

$(a_1 a_2 a_3 a_4 \ldots a_n)^{-1} = (a_1 a_n a_{n-1} \ldots a_3 a_2)$.

Now, does that make sense?