Set of generators for $A_n$, the alternating group. The problem is this: Prove that $A_n = \langle (123),(124),\ldots,(12n)\rangle$.
I had cogitated this problem for quite awhile, and haven't been able to come up with anything. 
The only good idea (at least I thought it was relatively good) that I had was to try to prove that the subgroup generated by these elements is a normal subgroup of $S_n$ which would force it to be $A_n$, but I guess that only works for $n \geq 5$.
 A: I know it's been a while, but may be useful for someone.
So, we have all of the $3$-cycles of form $(12x)$, also $(12x)(12x)$ is $(1x2)$, so we have them too.
As you can see, $(12y)(12x)(1y2)=(1xy)$, so we can get all the $3$-cycles with $1$ at the first place and anything elsewhere.
Next, $(1zx)(1xy)=(xyz)$, so we can get any $3$-cycle.
Now, any even permutation is a composition of permutations in forms $(ab)(cd)$ or $(ab)(ac)$, which we can easily get from $(acb)=(ab)(ac)$ and $(acb)(acd)=(ab)(cd)$
A: Given any distinct $a,b,c$, we want to express $(abc)$ as a product of $3$-cycles of the given form. Well, let's play around a bit. Multiplying two arbitrary $3$-cycles of the given form together yields:
$$
(12x)(12y) = (1x)(2y)
$$
Hmm, that didn't get us anywhere. In hindsight, if we want $a$, $b$, and $c$, then it makes sense that we'll need at least three $3$-cycles. Okay, so let's try that:
$$
(12a)(12b)(12c) = (1b2ca)
$$
Interesting. Now how many more $3$-cycles do we need to multiply by until we get what we want? Well, we could solve for it algebraically. Call the rest $\alpha$. Then:
$$
(1b2ca)\alpha = (abc) \implies \alpha = (1b2ca)^{-1}(abc) = (1ac2b)(abc) = (1a)(2b)
$$
Hold on a second. We got two $2$-cycles... that seems familiar. Can you see how to finish?

Remark: This only handles the case where $a,b,c \geq 3$. The $(12a)$ case is immediate. So there are two other cases to be handled separately: $(1ab)$ and $(2ab)$.
A: Hint.
You know (or you can prove) that the transpositions $(1i)$ generate $S_n$. As $A_n$ is the set of permutations written as an even number of transpositions, we just have to generate the products $(1i)(1j)$ of two transpositions. So let's do it...
For $i,j \ge 2$ and $i \neq j$ you have $(1i)(1j)=(1ji)$.
You also have for $i,j \ge 3$ and $i \neq j$ $$(1ij)=(12j)(12i)(12i)$$ Finally $(1i2)=(12i)(12i)$ for $i \ge 2$
All this will guide you to the conclusion.
