# 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 $n \geq 5$.

• With this set, can you generate all 3-cycles (think conjugation). Can all elements of $A_n$ be written as products of 3-cycles? Jul 12, 2015 at 1:10
• @ are you hinting that all elements can be written as products of 3-cycles or are you asking me if they can be? Jul 12, 2015 at 1:32
• Every even permutation is a product of permutations of either of two forms: $(ab)(cd)$ and $(ab)(bc)$. The latter form is a $3$-cycle, and every $3$-cycle is of that form. So it's enough to figure out how to write every permutation of either of those two forms as a product of permutations of the forms given in the question. One can show that every permutation of the form $(ab)(cd)$ is a product of two $3$-cycles (an exercise that probably won't take you very long). Now the problem is reduced to this: Can every $3$-cycle be written as a product of the given $3$-cycles? ${}\qquad{}$ Jul 12, 2015 at 1:50
• Your idea of showing that the subgroup is normal is great. The fact that it 'only' works for $n\geq5$ isn't really a problem; for $n\leq3$ the claim is obvious and for $n=4$ it is not hard to show directly. Mar 3, 2020 at 9:15

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 $$(acb)(acd)=(ab)(cd)$$

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 $(2ab)$.

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.

• how does this give conclusion? Jul 12, 2015 at 7:02
• @RipanSaha With what I wrote, any product $(1 \ i)(1 \ j)$ can be written as a product of $(1 \ 2 \ k)$. And $a_n$ is generated by such $(1 \ i)(1 \ j)$ products. Mar 3, 2020 at 12:29