What is the square of a cycle in S_7 symmetric group?

Suppose there's a cycle in where is a symmetric group on a set

What is ?

Is it composition of like this ?

I'm asking this because suppose there's a function

Now, whereas

which are completely two different things.

So what is the meaning of (square of a cycle) in abstract algebra?

Re: What is the square of a cycle in S_7 symmetric group?

yes, "what a square is" depends on "what your multiplication is".

with real-valued functions, there are indeed "two different multiplications" we might use: composition, or "point-wise multiplication" (the (x+1)^{2} one).

however, in permutation groups, the group operation is composition (permutations are nothing more than bjiective functions on finite sets), so α^{2} = αoα

so it makes sense, for example, to talk of α^{2}(4) = α(α(4)).

in a general group, "we don't know" what our "multiplication" is (although we do know for SPECIFIC groups). it's just a binary operation that satisfies some axioms. the word "multiplication" is sort of misleading, "group operation" is more accurate.

in your example, (3 7 1 4)^{2} is this function:

1-->4-->3

2-->2-->2

3-->7-->1

4-->3-->7

5-->5-->5

6-->6-->6

7-->1-->4

which is the permutation (1 3)(4 7) (it is customary to write (3 7 1 4) as (1 4 3 7), but not necessary).

there is a nifty trick for computing powers of a k-cycle:

for the square, "skip one space". so instead of sending 3 to 7, we skip one and send 3 to 1. we next do the same thing with 1.

for the cube, "skip two spaces".

for the n-th power "skip n-1 spaces".

so (1 4 3 7)^{3} sends 1 to 7 (1 7....

and then sends 7 to (skip 1, skip 4) uh..."3"! (1 7 3....)

and sends 3 to (skip 7, skip 1) 4, and sends 4 to (skip 3, skip 7) 1 (close cycle) (1 7 3 4).

Re: What is the square of a cycle in S_7 symmetric group?