1. ## groups

Thank you.

2. Can anyone help here?

Do I make a Cayley table? If so, Im confused how to traverse the permutation. If I start at (15)(24), it goes 1 to 5, 5 to 4, 4 to 1 ... is this correct for first step?

thanks

3. Originally Posted by theory07
If so, Im confused how to traverse the permutation. If I start at (15)(24), it goes 1 to 5, 5 to 4, 4 to 1 ... is this correct for first step?
No. Lets write in traditional notation.
$\displaystyle \left( {15} \right)\left( {24} \right) = \left( {\begin{array}{*{20}c} 1 & 2 & 3 & 4 & 5 & 6 \\ 5 & 4 & 3 & 2 & 1 & 6 \\ \end{array}} \right)$

1 to 5, 5 to 1; 2 to 4, 4 to 2. The 3 & 6 are inactive.

4. Is that why it makes a group, because whatever number you start from, you always get back to the same number.. so you just get e. ?

5. Originally Posted by theory07
Is that why it makes a group, because whatever number you start from, you always get back to the same number.. so you just get e. ?
I don't understand what you mean. But I will tell you this.
First, you should have known about cycle notation before even tackling this problem.
You also should how to show that a subset of a group is a subgroup of the group.
In order to show that a subset H of G is a subgroup you must show that the identity is in H and that H is closed with respect to the group operation and group inverse.
Constructing a group table is one possible way to proceed.