Question

question 2

Answer 1

alright question 8 where it says finding a subgroup H of S6

Answer 2

so i know D3 is generated by an element of order 3 and an element of order 2

Answer 3

so right now we are finding an element of order 3 and an element of order 2 in S6?

Answer 4

Right. I previously suggested using \((123)\) and \((12)\).

Answer 5

yeah that has order 3 and 2 in S6 but how are we showing that this is isomorphic to H

Answer 6

So using those two elements, look at the group generated by it.\[<(123),(12)>=\{(e,(123),(132),(12),(12)(123),(12)(132)\}\]Where \((123)\longmapsto r\) and \((12)\longmapsto s\). To show that this is an isomorphism, you need to verify two things.

1. That \(<(123),(12)>=\{(e,(123),(132),(12),(12)(123),(12)(132)\}\) is indeed an equality.

2. That \((12)(123)=(132)(12)\). This is showing \(sr=r^2s\).

Answer 7

ok so i am confused what are we verifying :s that its a equality ? :s

Answer 8

You need to verify that the equality here\[<(123),(12)>=\{(e,(123),(132),(12),(12)(123),(12)(132)\}\]is true. I didn't check it very carefully. Basically, just make sure that there aren't any extra elements that I missed.

Answer 9

alright yeah it holds

Answer 10

sorry i was just checking and for 2. we just state that fact ? :s

Answer 11

As long as you've checked that \((12)(123)=(132)(12)\), you can just state that since \(|(123)|=3\), \(|(12)|=2\), and \((12)(123)=(132)(12)\) it must be isomorphic to \(D_3\).

Answer 12

alright perfect awesome KingGeorge thanks a lot right now i am trying to learn a little about automorphism i shall ask question later on about the same questions

Answer 13

Good luck.