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If $H$ is a normal subgroup of $S_n$ where $(12)(34)∈ H$, $Sn/H \cong {e}$ or $Sn/H \cong Z/2Z$.

I need to prove the above statement and I have figured out that

  1. $Sn/H \cong {e}$ means that $H$ is $S_n$.
  2. $Sn/H \cong Z/2Z$ means that $H$ is $A_n$.

If I'm right, I need to prove that if $(12)(34)∈ H$, $H = A_n$ or $H = e$.

I can see that $H$ could be $A_n$ as $(12)(34)$ is even. However, isn't it possible that there exits some normal subgroup of $S_n$ that contains $(12)(34)$ but not $A_n$?

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Let us consider $S_4$, then it is easy to verify that $H=\{e, (1\,2)(3\,4), (1\,3)(2\,4), (1\,4)(2\,3)\}$ is a normal subgroup of $S_4$ because $H$ is the union of conjugacy classes of $e$ and $(1\,2)(3\,4)$. Moreover $H$ contains the element $\color{red}{(1\,2)(3\,4)}$ but this subgroup $H$ is neither $A_4$ nor $S_4$.

So the statement you have in the title is false for $n=4$ as $\left|S_4/H\right|=6$.


Added text: However it can be shown that for $n \geq 5$ the only non-trivial normal subgroup of $S_n$ is $A_n$ (see Proving that $A_n$ is the only proper nontrivial normal subgroup of $S_n$, $n\geq 5$).

Thus if $(1 \,2 )(3 \, 4) \in H$ and $H$ is normal in $S_n$, then $H=S_n$ or $H=A_n$ and the statement follows.

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  • $\begingroup$ would the statement be true for $n>4$? $\endgroup$ Jul 8, 2019 at 8:00
  • $\begingroup$ @J.W.Tanner Yes it would be true. I have added some stuff to my answer above. $\endgroup$
    – Anurag A
    Jul 8, 2019 at 8:09
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    $\begingroup$ Thank you. +1. As you may see, I anticipated that in my comment to the other answer $\endgroup$ Jul 8, 2019 at 8:11
  • $\begingroup$ @J.W.Tanner You intuition was right on the spot :-) $\endgroup$
    – Anurag A
    Jul 8, 2019 at 8:12
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Hint: Take the sign mapping $S_n\rightarrow\{\pm 1\}:\pi\mapsto {\rm sgn}(\pi)$. For $n\geq 2$, the mapping is an epimorphism and the kernel (normal subgroup) is the alternating group $A_n$. Since the permutation $(12)(34)$ is even, it lies in the kernel. The multiplicative group $\{\pm1\}$ is isomorphic to the additive group $Z/2Z$. By hypothesis, the kernel $H$ has order $n!/2$.

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    $\begingroup$ so $(12)(34)\in A_n$, but can you explain (as OP asked) how we'd know there isn't a normal subgroup of $S_n$ that contains $(12)(34)$ and is a proper subset of $A_n$? do we need to know that $A_n$ is simple for $n\ge5$? $\endgroup$ Jul 8, 2019 at 7:52
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If $n\ge5$, then $(123)=(12)(23)=(12)(45)\cdot(45)(23)$.
Since conjugates of $(12)(34)$ are all $(ab)(cd)$ for distinct $a, b, c, d$, we get all 3-cycles in $H$, but these generate $A_n$.

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  • $\begingroup$ This post supports your assertion that $3$-cycles generate $A_n$ $\endgroup$ Jul 8, 2019 at 8:24

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