Generators of $A_5$

Question

I read somewhere that $(123), (124),$ and $(125)$ generates $A_5$; I checked that but I am looking for a good proof.

I am also wondering if there is a generalization for this fact for example $\langle(123),(124),...,(12n)\rangle\stackrel{?}{=}A_n$.

Answer 1

Knowing that who can do more can do less... answering your second question will also answer the first one. So let's be lazzy and only answer the second one!

Name $\langle(123),(124),...,(12n)\rangle= A_n^\prime$. You indeed have $A_n = A^\prime_n$. The inclusion $A^\prime_n \subseteq A_n$ is trivial. So let's prove converse inclusion.

Take $\sigma \in A_n$. $\sigma$ can be written as $$\sigma= t_1 t_2 \dots t_{2p} = t_1 t_0 t_0 t_2 t_3 t_0 t_0\cdots t_0 t_{2p}=(t_1 t_0)(t_0 t_2)(t_3 t_0) \dots (t_{2p-1} t_0)(t_0 t_{2p})$$ where $t_1, t_2,\dots, t_{2p}$ is an even number of transpositions and $t_0 = (1\, 2)$.

Now you can prove that for $i,j \notin \{1,2\}$: $$[(1 \, 2 \, i), (1 \, 2 \, j)] = (1\, 2)(i \, j) = (i\, j)(1 \, 2) \in A^\prime_n$$

while for $i \notin \{1,2\}$: $$\begin{cases} (1\, 2)(1 \, i) = (1 \, 2 \, i)^2 \in A^\prime_n \text{ and } (1\, i)(1 \, 2) = (1 \, 2 \, i) \in A^\prime_n\\ (1\, 2)(2 \, i) = (1 \, 2 \, i) \in A^\prime_n \text{ and } (2\, i)(1 \, 2) = (1 \, 2 \, i)^2 \in A^\prime_n \end{cases} $$

Gathering all the above, you're done. For the reminder, $[x, y]=xyx^{-1}y^{-1}$ is the commutator of $(x,y)$.

Answer 2

The group generated by just two of your generators obviously generates a copy of $A_4$ (argue by order and Lagrange's theorem) and doesn't contain the other generator. If you then add the other generator, by Lagrange, the generated group must be of an order that divides 60 and is divisible by (but not equal to) 12. since $60/12$ is prime, you must have generated all of $A_5$.

You can in fact do better for $A_{2k+3}$: It's generated by, e.g. $(123),(345),...,(2k+1\, 2k+2\, 2k+3)$, and if you want $A_n$ for even $n$, add a $(2k+2\,2k+3\,2k+4)$. So for example, $A_5$ is generated by just $(123),(345)$. To see that, check that their product is a $5$-cycle.