# Is there a more elegant way of proving $\langle (1,2)(3,4), (1,2,3,4,5) \rangle = A_5$

I'm trying to show the following

$\langle (1,2)(3,4), (1,2,3,4,5) \rangle = A_5$

I managed to prove this but I think my solution is very inelegant. Here's my argument

let $J = \langle (1,2)(3,4), (1,2,3,4,5) \rangle,$ then $\lvert J \rvert \geq 8$ and $\lvert J \rvert$ divides $\lvert A_5 \rvert = 60,$ so the possibilities are

$\lvert J \rvert = 10, 12, 15, 20, 30, 60$.

Then I sat down and calculated 13 more elements of $J$, so now the possibilities are $\lvert J \rvert = 30$ or $60$. But we can't have $\lvert J \rvert = 30$ because then $J$ would be normal (index 2 theorem) which would contradict $A_5$ being simple, so we must have $J = A_5$.

The calculation part make this proof quite long winded, is there a simpler way of getting the result?

• Another approach, and an interesting one, is to show that with those generators one is able to produce all the cycles of length three, or 3-cycles, of $S_5$. Now it follows that it generates $A_5$, as $A_n$ is generated by all the 3-cycles of $S_n$. Dec 23, 2014 at 12:16

As a more elementary approach $H=<(1 2)(3 4),(1 2 3 4 5)>$ contains an element of order 2, and element of order 5 and an element of order 3 (the product of the two generators).

• could you please elaborate more? I know that it containing an element of order X implies that X divides $|<(1,2)(3,4),(1,2,3,4,5)>|$. But having order 2,3, and 5 only means that 30 divides the number of elements in it? That's not sufficient to show that it has 60 elements and hence is $A_5$, right?
– jh4
Jan 17, 2015 at 19:44
• all the three cycles in $A_n$ generates it Jan 18, 2015 at 4:19

By conjugation, in the subgroup generated by $\tau=(1\,2)(3\,4)$ and $\gamma=(1\,2\,3\,4\,5)$ there is any double transposition of the form $(a,a+1)(a+2,a+3)$, so there is $\sigma= (1 \,2)(4\,5)$, and by conjugation again, there is $(1\,2)(3\,5)$, so there is any double transposition exchanging two consecutive elements ($5$ and $1$ are considered consecutive). Acting by conjugation again, in the generated subgroup there is any double transposition, so any $3$-cycle, so $A_5$.

A5 is generated by a 3 cycle. The product of the generators is 3 cycle so it generates A5

• No, A5 is not generated by a 3-cycle. A 3-cycle generates a cyclic group of order three. Jan 18, 2015 at 4:06
• Why not @Maiano Suarez-Alvarez ?let A∈ An then is A can be expressed as the product of even number of transposition. And also (a,b)=(1.a)(1.b)(1.a).THen A is a product of 3 cycle. Arbitrariness of A does the proof. Am i not right? Jan 18, 2015 at 4:13
• @SayantanKoley you are way beyond wrong.... Jan 18, 2015 at 4:21
• Correct me plz. Where is my fault ? @BhaskarVashishth Jan 18, 2015 at 4:45
• What order does a 3 cycle have? It is order 3. What order does $A_5$ have? 60. Just a little bit off. Jan 18, 2015 at 6:23