Jamie Mulholland p. 69 Theorem 6.1 or Fraleigh p. 90 Corollary 9.12
Any permutation of a finite set of at least two elements is a product of 2-cycles.
$1. (a_1, a_2, ···,a_n)= (a_1, a_n)(a_1, a_{n-1}) · · · (a_1, a_3)(a_1, a_2). \implies 5. (a, b)(a, c) = (a, b, c) $
Jamie Mulholland p. 75 for 2 to 4 and p. 86 for 5, 6. I skip $(c,d)(a,b) = (a,b)(c,d)$.
$2.$ $(a, b)(a, b) = id \qquad 3. (a,c)(a, b) = (a, b)(b, c) \qquad 4. (b, c)(a, b) = (a, c)(c, b)$
$6. (a, b)(c, d) = (a, b, c)(a, d, c)$
I tried. I know how to check LHS = RHS by following each element but I don't want to do this every time. What's the intuition for the 6 identites?
You can visualize the LHS as moving every letter to its destination all at once, while the RHS moves each letter one at a time, keeping $a_1$ as a "working" slot until the end.
For example, let's look at $\sigma \triangleq (1,2,3,\cdots, n)$. You can envision picking all the numbers up together, shifting them over by one slot, and then sitting them back down again. That's the LHS. On the RHS, you just swap two at a time. Each time, you take whatever is in the $a_1$ spot to the spot that $\sigma$ is supposed to put it, and bring whichever letter is in that spot back with you to the $a_1$ spot. When you finally get to $a_n$ after moving all the other letters, every letter will have been moved to its proper place, including $a_n$, which belongs in the $a_1$ spot.
If a coin faces heads up and you turn it over twice, you'll end up back where you started, right? In this example, heads is $a$, and tails is $b$.
Using #2, we can multiply on the left by $(a,b)$ to get $(a,b)(a,c)(a, b) = (b, c)$. By #1, the LHS is $(b,c,b)=(b,c)$.
This is the same thing as #3. Since $(b,c)=(c,b)$ and $(a,c)=(c,a)$, we can multiply on the left by $(c,b)$ to get $(a, b) = (c,b)(c,a)(c, b)=(b,a,b)=(a,b)$.
This is a special case of #1.
My best advice for this one is to just work it out on a table with some cups, cards, or coins. Pretend you're a card dealer- permute! Get a feel for it.