Prove: The symmetric group $\mathfrak{S}_4$ does not have a normal subgroup of order 2.

I found the following way of showing this, which does not make me happy. My questions are: Is this proof correct? And: Is there a nicer way of showing this, by using the more abstract concepts of algebra, for example isomorphism theorems and sylow theorems? For example, how could I determine, whether there is a normal subgroup of order 4?

My attempt: Suppose there is a normal subgroup $S \unlhd \mathfrak{S}_4$. For satisfying the group axioms, $S$ must have the structure $S = \{e, g\}$, where $e \in \mathfrak{S}_4$ is the neutral permutation and $g \in \mathfrak{S}_4$ any selfinverse element, that means $gg = e$. For permutations, this means, that $g$ has cycles of length 1 and 2 only, but at least one cycle of length 2, since it is not the neutral permutation. Now I can inspect the two possible cases:

Case 1: g consist of two cycles of length 2

Case 2: g constst of one cycle of length 2 and two cycles of length 1

For both cases, it is easy to find a $h \in \mathfrak{S}_4$ so that $hgh^{-1} \neq g$ (I don't want to step into messy details here). But it is also $hgh^{-1} \neq e$, since $hgh^{-1} = e$ would lead to $g=e$. So we have $hgh^{-1} \notin S$, which shows that $S$ is not a normal subgroup.

Thank you for your help.

  • 2
    $\begingroup$ the proof is correct $\endgroup$ – Jorge Fernández Hidalgo Aug 21 '16 at 22:44
  • 2
    $\begingroup$ Shouldn't it be $hgh^{-1}\neq g$ instead of $hgh^{-1}\neq h$? $\endgroup$ – Arthur Aug 21 '16 at 22:47
  • $\begingroup$ @Arthur: This is right, thank you. It was just a typing mistake. I've corrected it. $\endgroup$ – S. M. Roch Aug 21 '16 at 23:22

Two elements of $S_n$ are conjugate if and only if their cycle decompositions have the "same shape".

To see this, first recall the beautiful way that conjugation works in $S_n$:

If $\tau = (abcd \dots)$ is a cycle, and $\sigma$ is a permutation, then $\sigma \tau \sigma^{-1}$ is the cycle $(\sigma(a) \sigma(b) \sigma(c) \sigma(d) \dots)$, i.e. conjugation by $\sigma$ precisely relabels the elements in the cycle $\tau$ according to the permutation $\sigma$!

(I always found this really cool. It's analogous to the way that if you change basis in a vector space by a matrix $A$, and if $L$ is a linear transformation given by the matrix $B$ w.r.t. the original basis, then w.r.t. the new basis $L$ is given by the conjugate matrix $A B A^{-1}$.)

Using this, you can see that if you have two cycle decompositions with the same shape (i.e. involving the same numbers of cycles of a given length), then you can just write down a permutation $\sigma$ which conjugates one to the other.

Now use the fact that any normal subgroup of a group $G$ contains the entire conjugacy class (and here I mean the conjugacy class in $G$) of any of its elements. (This is equivalent to normality, as you can easily check.)

So if $H \leq S_n$ is normal and contains an element $\tau$ of order $2$, it contains all the elements that have the same cycle shape as $\tau$.

E.g. if $\tau$ is a $2$-cycle, well there are six $2$-cycles altogether, and so $H$ has to contain all six of them (at least). In particular, it can't have order $2$.

E.g. if $\tau$ is a product of disjoint $2$-cycles, there are three of those, and so $H$ has to contain all three of them, and also the identity, so has order at least $4$. Now you can check: the subset consisting of the identity and these three products of disjoint $2$-cycles is a union of conjugacy classes, and so is closed under conjugation. Is it closed under multiplication? If it is, you've found a normal subgroup of order $4$.

  • $\begingroup$ The "shape" of the cycle decomposition of a permutation is called the cycle type of the permutation. $\endgroup$ – Pedro Tamaroff Aug 22 '16 at 0:54
  • $\begingroup$ Thank you for this useful answer. It is exactly what I have wished. $\endgroup$ – S. M. Roch Aug 22 '16 at 1:16
  • $\begingroup$ @S.M.Roch You're welcome! $\endgroup$ – tracing Aug 22 '16 at 1:45

Every subgroup of order two in $S_4$ is generated by either a transposition $\sigma = (ij)$ or a double transposition $\rho = (ij)(kl)$. If $\tau = (il)$ then

$$\tau\sigma\tau^{-1} = (jl)$$ $$\tau\rho\tau^{-1} = (lj)(ki)$$

are not in this subgroup of order two. It seems a bit unsatisfactory to claim that "there is $h$ such that $hgh^{-1}\neq g$" without proof, because that's the crux of the argument. You must exhibit such $h$.

  • $\begingroup$ This is right. Thank you. But it's still the same idea, to use the elementar properties of permutations. $\endgroup$ – S. M. Roch Aug 21 '16 at 23:25

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