Find the number of elements of order 3 in $S_7$

Question

I understand that there are two cycles of length 3, $(i,j,k)(a,b,c) \in$ $S_7$.

However, I'm quite stumped in figuring out the logic behind these steps, leading to the answer :

Number of distinct 3 cycles : $(7*6*5)/3$ = 70

Number of distinct elements of the form (i,j,k)(a,b,c) = $(70*((4*3*2)/4))/2$ = 280 distinct elements

I would appreciate any help in terms of understanding the reasoning behind these steps.

Answer 1

There are the 3-cycles:

For each choice of $\{a,b,c\}$ there are two 3-cycles: $(a,b,c)$ and $(a,c,b)$. The total number of them is: $$ 2\times \binom 73 = \frac{7\times 6\times 5}{3} = 70 $$

Then when there are two 3-cycles: first choose the elements of the first 3-cycle, then there are 2 3-cycles possible. Then choose the 3 elements of the second 3 cycles, then the 3-cycle.

Eventually you count every element of order 3 twice (the first and second cycle being both 3-cycle): so divide the final number by 2. $$ \frac 12\times \left[2\times\binom 73\right]\times \left[2\times \binom 43 \right]= 70\times 4 = 280 $$

so there is a total of $350$ elements of order 3.