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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?

user78655
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  • 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 $S_n$. – Diego Robayo Dec 23 '14 at 12:16

3 Answers3

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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).

John McGee
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  • 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 '15 at 19:44
  • all the three cycles in $A_n$ generates it – Bhaskar Vashishth Jan 18 '15 at 04:19
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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 $A_5$.

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

Sayantan Koley
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    No, A5 is not generated by a 3-cycle. A 3-cycle generates a cyclic group of order three. – Mariano Suárez-Álvarez Jan 18 '15 at 04:06
  • 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? – Sayantan Koley Jan 18 '15 at 04:13
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    @SayantanKoley you are way beyond wrong.... – Bhaskar Vashishth Jan 18 '15 at 04:21
  • Correct me plz. Where is my fault ? @BhaskarVashishth – Sayantan Koley Jan 18 '15 at 04:45
  • What order does a 3 cycle have? It is order 3. What order does $A_5$ have? 60. Just a little bit off. – Vladhagen Jan 18 '15 at 06:23
  • Your comment and your answer seem to be at odds with each other. When you write $(a,b) = (1,a)(1,b)(1,a)$, you are not writing $(a,b)$ as a 3-cycle, but rather writing $(a,b)$ as a product of three transpositions (2-cycles). – user642796 Jan 18 '15 at 07:28
  • Then what i meant that A=σ1σ2....*σr, r is even with σi is transpositions for all i=1,2,...,r. Each σi=(a,b)=(1,a)(1,b)(1,a) then A=(1,i1)(1,i2)...(1,im) where m is even in fact m=3r. Now (1,i1)(1,i2)=(1,i2,i1). Am i not right @ArthurFischer♦ – Sayantan Koley Jan 18 '15 at 12:51