How do I find the subgroups of $S4$ generated by these sets in each case?
- $A = {(1,3),(1,2,3,4)}$
- $B = {(1,2,4),(2,3,4)}$
- $C = {(1,2),(1,3),(1,4)}$
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Martín Mas
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Do you know what "generate" means? – Secret Math Oct 17 '13 at 18:10
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By doing the computations. There are not actually that many of them to do. – Tobias Kildetoft Oct 17 '13 at 18:24
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Thanks to Babak S. for the answer. Also, thanks to Secret Math and Tobias Kildetoft for the rude comments. – Martín Mas Oct 17 '13 at 19:15
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@MartínMas: the first example shows that the claim "a subgroup of $S_n$ generated by a $n$-cycle and a transposition is $S_n$ itself" is wrong if $n$ is not prime. – Watson Sep 07 '16 at 12:25
1 Answers
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I am giving you some points about A. I hope you can use them well. Let $a=(1,3),~b=(1,2,3,4)$ , we know that $$S_4=\langle x,y\mid x^2=y^4=(xy)^3=1\rangle$$ and beause of the final relation inside the presentation, we can show every element of $S_4$ as $x^iy^j$ where $i=0,1,~~j=0,1,2,3$. Here, $|a|=2,~~|b|=4$ and $$a^2=id\\ b^2=(1,3)(2,4)\\ b^3=(1,4,3,2)\\ab^2=(2,4)\\ab^3=(1,2)(3,4)\\ ab=(1,4)(2,3)$$ This means that $\langle a,b\rangle=\{id,a,b,ab,ab^2,ab^3,b^2,b^3\}$
Mikasa
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I'm not pretty sure about how to do it with $B$. Do I have to compute $a^2$, $b^2$, $ab$, $a^{2} b$ and $ab^{2}$ ? Why do I haven't got to compute $ba$, $b a^{2}$ and $b a^2$?
Also, is $C = S_4$?
– Martín Mas Oct 20 '13 at 17:17 -