Hi I have an exam very close, and I would appreciate if anyone could answer the queries that I have for these questions.
I particularly need clarifcation for iii v and vi.
The general question that I have is to:
Perform the following calculations of elements of S5 in cycle notation:
(i) (21534)^-1
(ii) (5124)^2
(iii) (45).(34).(45)
(iv) (12)^(45)
v) ((154)(23))^7
vi) ((12).(23).(34).(45))^5
So for:
i) Quite straightforward, just flip the bracket round so that it reads: (43512).
ii) Again easy. (5124)^2 = (5124)(5124). Evaluating this has: (52)(14)
iii) I have reservations about the . used here, how does the dot differ as to if it was expressed as: (45)(34)(45) Would there be any difference? I have the final answer being (4)(53) so (53).
iv) I am not completely familiar with these types. If it is of the form A^B, then this is equivalent to B^-1 AB right? So (12)^(45) becomes (54)(12)(45) which is simply (12)
v) Since the elements within the brackets are not repeated I have this as (154)(23).
vi) Although I consider that I have the correct answer as (45123) I am unsure of how this is really achieved. The power is negligent in most cases, in cycle notation it goes from right to left:
So by starting at 4, this goes to 5. But do we stop there or go all the way back around to get 1. Then stop - or go all the way back around to get 2 ergo 3.
Any help would be so appreciated. Thanks
If I have posted this incorrectly or tagged it wrong, I'm sorry. I am relatively new to this forum.
I'm just going to help you with (vi), which I believe will be the most useful for you.
To evaluate:
$[(1\ 2)(2\ 3)(3\ 4)(4\ 5)]^5$, let's evaluate what's "inside" the brackets, first.
Working right-to-left, we see that:
$1 \to 1 \to 1 \to 1 \to 2$ (only the last transposition (on the left) moves $1$).
$2 \to 2 \to 2 \to 3 \to 3$ (the penultimate transposition moves $2$, but the last one doesn't move $3$).
$3 \to 3 \to 4 \to 4 \to 4$
$4 \to 5 \to 5 \to 5 \to 5$
$5 \to 4 \to 3 \to 2 \to 1.$
We can re-write this as $(1\ 2\ 3\ 4\ 5)$, a $5$-cycle, so:
$[(1\ 2)(2\ 3)(3\ 4)(4\ 5)]^5 = (1\ 2\ 3\ 4\ 5)^5 = e$.