Thread: groups

Thank you.

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

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.


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

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. ?

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.