jango_IN_DTOWN one year ago Subgroup + permutation problem

1. jango_IN_DTOWN

2. jango_IN_DTOWN

@ganeshie8

3. ganeshie8

Hey still here ?

4. jango_IN_DTOWN

hii

5. ganeshie8

familiar with the cycle notation ? |dw:1444581237571:dw|

6. jango_IN_DTOWN

yes |dw:1444581322888:dw|

7. ganeshie8

Ah, so is it given that $$(1,2)\in S_3$$ ?

8. ganeshie8

|dw:1444581438396:dw|

9. jango_IN_DTOWN

no i assumed it to be S3 since the next two contains 3, we are not given about the symmetry group

10. ganeshie8

since $$\alpha$$ exists in the subgroup, $$\alpha^2, \alpha^3, \ldots$$ must also exist in the group. $$\alpha^2 = ?$$

11. jango_IN_DTOWN

|dw:1444581538927:dw|

12. jango_IN_DTOWN

the identity permutation

13. ganeshie8

so we have $$\alpha^2 = e$$, therefore the subgroup generated by $$(1,2)$$ is : $$\{e,\alpha\}$$

14. jango_IN_DTOWN

hmm true...

15. ganeshie8

you may express the same in cycle notation as $$\{(1), (1,2)\}$$

16. ganeshie8

try working ii similarly, just consider all the powers of given permutation

17. jango_IN_DTOWN

ok thanks...

18. jango_IN_DTOWN

we will stop when the identity permutation is reached

19. ganeshie8

elements will be repeated if you continue, notice that $$\alpha^n =e \implies \alpha^{n+1}=\alpha$$

20. jango_IN_DTOWN

Yes I got it.. Thanks..

21. ganeshie8

np