If N is a subgroup of $S_4$ and |N|=12 , N is normal in $S_4$ because N has index 2. N has also an element of order 3 for the Cauchy's lemma.Why N has all the 3-cycles of $S_4$?
I want to prove $S_4$ has only a subgroups of order 12
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Yes? Was that actually your full question? – Tobias Kildetoft Jan 29 '16 at 09:49
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do you want to prove that there is only one such subgroup? if yes, see this question: http://math.stackexchange.com/questions/27024/a-n-is-the-only-subgroup-of-s-n-of-index-2 – user8268 Jan 29 '16 at 09:53
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the professor gave us that tip to follow – Giulia B. Jan 29 '16 at 10:02
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All $3$-cycles are conjugate. If $N$ is normal and contains a $3$-cycle, then $N$ contains all $3$-cycles. It is well known that $A_n$ is generated by the $3$-cycles for $n\ge 4$. This means that $N \supseteq A_4$ and so $N = A_4$.
lhf
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Here is another way to look at it:
Since $N$ is normal, and it has an element of order $3$, it contains a $3$-cycle, and thus every $3$-cycle. For example, if $(1\ 2\ 3) \in N$, then taking $g$ to be the map:
$g(1) = a\\g(2)= b\\g(3) = c$
(and sending $g(4)$ to "whatever's left")
we have $g(1\ 2\ 3)g^{-1} = (a\ b\ c) \in N$.
So $N$ contains at least $9$ even permutations (the $3$-cycles and the identity), that is:
$9 \leq|N \cap A_4| \leq 12 = |N|$.
But $N \cap A_4$ is surely a subgroup of $S_4$, and thus has order dividing $24$.
Of the numbers $\{9,10,11,12\}$ only $12|24$, so we conclude $|N \cap A_4| = 12$, that is: $N \subseteq A_4$.
But the only subgroup of $A_4$ that has order $12$ is $A_4$ itself, so $N = A_4$.
David Wheeler
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