Prove that there doesn't exist any normal subgroup $H$ such that $S_5/H $ is isomorphic to $S_4$

Question

My attempt : Order of the group $S_5$ is $5!$ so by Lagrange's theorem the order of the group $H$ should be $5$.So it must be a cyclic group generated by a $5$ cycle then let $(a_1 a_2 a_3 a_4 a_5)$ be a generator of the subgroup. Then for any element $g \in S_5$ ,$g(a_1..a_5)g^{-1} \in H$ , then $g(a_1)..g(a_5) \in H$.Now, all the elements of $H$ are $5$ cycles. Now if we choose $g$ in such a manner that $g(a_1)..g(a_5)$ becomes a two cycle then I am done. So I choose $g(a_1)=a_2$,$g(a_2)=(a_1)$..Is this OK? I don't think it is right where am I going wrong?

Answer 1

You are on the right track, you've shown that such a normal subgroup would need to be generated by a $5$ cycle in $S_5$, so now we need to show that the subgroup generated by a $5$ cycle in $S_5$ can't be normal. Conjugating a $5$ cycle will always yield a $5$ cycle (since conjugation preserves the order of elements of a group), so we'll need to work a little harder. There are a few ways to see this, but I'll give a hands on proof. Its enough to show that from any $5$ cycle, we can find something in the subgroup generated by its conjugates that isn't a $5$ cycle, since then we can't have any group generated by a $5$ cycle being normal. So if we have $(abcde)$ is our $5$ cycle, conjugating by the transposition $(ab)$ yields the $5$ cycle $(bacde)$, and $(abcde)(bacde)^{-1}=(cba)$, which has order $3$, so gives something bigger than our hypothetical normal subgroup of order $5$. Thus, no normal subgroup of order $5$ can exist.

Answer 2

You have almost finished the proof. So $|H|=5$, $H$ consists of 4 5-cycles and the identity element. Since there are $24$ cycles of length $5$ in $S_5$ there exists a $5$-cycle $c$ not in $H$. Since any two $5$-cycles are conjugate $S_5$, there is a conjugation taking a cycle from $H$ to $c$. This is a contradiction since $H$ is normal.

Answer 3

You're doing well. Up to a relabeling of the elements which $S_5$ operates on, the generator of $H$ can be chosen to be $(12345)$. Now observe $$ (12)(12345)(12)=(13452)\notin H $$ and you're done.

Answer 4

We can use the fact that $A_5$ is simple:

Firstly, observe that if $N\leq H\leq G$ such that $N\not\lhd H$ then $N\not\lhd G$ (as if there exists some $h\in H$ such that $h^{-1}Nh\neq N$ there there also exists some $g\in G$ with this property - take $g:=h$).

As you noted, $H$ must be generated by a $5$-cycle $\sigma$. As $5$ is odd, $\sigma\in A_5$. As $A_5$ is simple, $\langle \sigma\rangle\not\lhd A_5$ and so $\langle \sigma\rangle\not\lhd S_5$ (using the above observation), as required.

This proof easily generalises to prove that $A_n$ does not surject onto $A_{n-1}$ for certain $n\geq5$. In increasing levels of difficulty:

1. If $n$ is prime. Here the proof works verbatim (prime is used to get the long-cycle $\sigma$).
2. If $n$ is odd. Here your kernel has odd order (as opposed to being a long-cycle), and so is wholly contained in $A_n$.
3. If $n$ is not a power of two (thanks to Aaron in the comments for this generalisation). Here the kernel $K$ contains an element $g$ such that $g^2\neq1$, and so $g^2\in A_n$ and $K\cap A_n\neq \{1\}$. Then, by properties of intersections of normal subgroups, $(K\cap A_n)\lhd S_n$. As $n>2$, $K$ has index greater than the index of $A_n$, so $K\cap A_n\lneq A_n$ and so $(K\cap A_n)\not\lhd A_n$ and so $(K\cap A_n)\not\lhd S_n$, a contradiction.

Answer 5

Suppose that such a $H$ exist. Then (First Homomorphism Theorem) there is a surjective homomorphism $\varphi\colon S_5\to S_4$ such that $\operatorname{ker}\varphi=H$. Now, $\forall \sigma,\tau\in S_5$ we get $\varphi(\sigma\tau\sigma^{-1})=\varphi(\sigma)\varphi(\tau)\varphi(\sigma^{-1})=\varphi(\sigma)\varphi(\tau)\varphi(\sigma)^{-1}\in \operatorname{Cl}(\varphi(\tau))$, whence:

$$\varphi(\operatorname{Cl}(\tau))\subseteq\operatorname{Cl}(\varphi(\tau)), \space\space\forall \tau\in S_5 \tag 1$$

Also, by surjectivity, $\forall s\in S_4, \exists \sigma\in S_5$ such that $s=\varphi(\sigma)$, and hence $\forall\tau\in S_5,\forall s\in S_4$ we get $s\varphi(\tau)s^{-1}=\varphi(\sigma)\varphi(\tau)\varphi(\sigma)^{-1}=\varphi(\sigma)\varphi(\tau)\varphi(\sigma^{-1})=\varphi(\sigma\tau\sigma^{-1})\in \varphi(\operatorname{Cl}(\tau))$, and hence:

$$\operatorname{Cl}(\varphi(\tau))\subseteq \varphi(\operatorname{Cl}(\tau)), \space\space\forall \tau\in S_5 \tag 2$$

By $(1)$ and $(2)$:

$$\varphi(\operatorname{Cl}(\tau))=\operatorname{Cl}(\varphi(\tau)), \space\space\forall \tau\in S_5 \tag 3$$

thus, say, $\varphi$ maps each conjugacy class to the conjugacy class of the homomorphic elements of the former.

Since $H\le S_5$ is cyclic of prime order ($5$), all its four non-identity elements have order $5$, i.e. are $5$-cycles, and hence they are all elements of one same conjugacy class of cardinality $24$, namely $\operatorname{Cl}((12345))$:

$$H\setminus\{Id_{S_5}\}\subsetneq \operatorname{Cl}((12345)) \tag 4$$

Take $\eta\in H\setminus\{Id_5\}$ and $\rho\in\operatorname{Cl}((12345))\setminus H$ (such a $\rho$ exists by $(4)$); then, by $(4)$ and $(3)$, $\varphi(\eta)\in \operatorname{Cl}(\varphi(\rho))$; but $\eta\in H=\operatorname{ker}\varphi$, so $\varphi(\eta)=Id_{S_4}\notin \operatorname{Cl}(\varphi(\rho))$, because $\rho\notin H$: contradiction, and such a $H$ doesn't exist.