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

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 $$A_n$$?

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.

• would the statement be true for $n>4$? Jul 8, 2019 at 8:00
• @J.W.Tanner Yes it would be true. I have added some stuff to my answer above. Jul 8, 2019 at 8:09
• Thank you. +1. As you may see, I anticipated that in my comment to the other answer Jul 8, 2019 at 8:11
• @J.W.Tanner You intuition was right on the spot :-) Jul 8, 2019 at 8:12

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 $$n!/2$$.

• 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 $n\ge5$? Jul 8, 2019 at 7:52

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 $$A_n$$.

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