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I was reading this question :

Normal subgroups of $S_4$

And I had a difficulty in understanding this statement:

"Now, any subgroup that contains all transpositions is the whole group."

Could anyone explain for me why this statement is true please?

2 Answers2

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The transpositions generate the symmetric group: each permutation is a product of transpositions. For example, a cycle $$(a_1\ a_2\ \cdots\ a_k)=(a_1\ a_2)(a_2\ a_3)\cdots(a_{k-1}\ a_k).$$

Angina Seng
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  • so he mean that any subgroup which contain $(12)(34)$ will generate the whole group and so it will be considered as a trivial normal subgroup? –  Feb 19 '19 at 19:49
  • all transpositions here mean $(12)(34)$? –  Feb 19 '19 at 19:50
  • but by this we will have 15 elements not 12 as he mentioned –  Feb 19 '19 at 19:52
  • "So we can consider only subgroups that don't contain the transpositions. Their order must be a divisor of 24, and since it does not have the transpositions, it is at most 12." this was a statement there also. –  Feb 19 '19 at 19:54
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    There are $6$ transpositions, but $(12)(34)$ is not a transposition. It's the product of two separate transpositions, $(12)$ and $(34)$. As for your last comment, since there's some element that's not in the subgroup, its order must be a divisor of $24$ that's less than $24$, which means it can't be any bigger than $12$. – Robert Shore Feb 19 '19 at 19:59
  • why there are some elements that is not in the subgroup?@RobertShore –  Feb 19 '19 at 20:25
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The answer to the linked question states that if a subgroup $H$ of $S_4$ contains all transpositions $(i\,j)$ for $i=1,2,3,4$, $j\ne i$ (there are $6$ such permutations because $(i\,j)=(j\,i)$), then $H=S_4$. It follows from the fact that any element (permutation) of the symmetric group $S_4$ (and more generaly $S_n$) is composition of a finite number of transpositions; in other words, transpositions of $S_4$ form a system of generators for $S_4$: explicitly $$\langle (1\,2),(1\,3),(1\,4),(2\,3),(2\,4),(3\,4)\rangle =S_4$$.

LBJFS
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