Cycle Notation Calculations of S6

Question

I particularly need help on i and iii

The general question that I have is to:

Perform the following calculations of elements of S6 in cycle notation:

i) ((156)(234))^-1

ii) (134).(12)(45)

iii) (12)(45).(134)

iv) (123)(45).(45)

v) (1234)^(15)(24)(36)

vi) (12)(34).(13)(24).(14)(23)

i) ((156)(234))^-1 = (156)^-1 (234)^-1 = (651)(432)

ii) (134).(12)(45) What does the dot mean does it have a different effect to having no dots as the second brackets do.

Anyway I have (51234).

iii) I have this as (135) I am unsure if this is correct?

iv) I have (123)(4)(5) reducing to (123) I am fairly sure this is correct.

v) Applying A^B = B^-1 AB = (3)(6254) which reduces to (6254).

vi) Is fairly simple - (2)(3)(1)(4) reduces to e.

Any help would be so appreciated thnx

Answer 1

i) ii) I have the same results

iii) I have $(13542)$

iv) yes, first two cycles are inverses, and cancel each other out.

v) $(1234)^{(15)} = (15)(1234)(15) = (5234)\\ (5234)(24)(36) = (25)(36)$

if it is supposed to be:

$(1234)^{(15)(24)(36)} = (15)(24)(36)(1234)(15)(24)(36)$

I think of conjugation as a change of bases. I doesn't change the structure of the cycles. It does change the way elements map in a rather straight forward way. $(5462)$

vi) $(12)(34),(13)(24),$ and $(14)(23)$ generate a group isomorphic to the $K_4$ group. Their product equals the identity.