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I know that we can show $A_n \trianglelefteq S_n$ by observing that:

  • $A_n$ is the set of all permutations in $S_n$ which are a product of an even number of transpositions. Hence any element is the product of an even number of even-length cycles.
  • Conjugation by an element of $S_n$ preserves cycle structure. (i.e. conjugation by an element of $S_n$ is just "relabeling" the elements).

Then we have that for all $a \in A_n$, for all $s \in S_n$, conjugating lands you back in $A_n$, i.e. $sas^{-1} \in A_n$.

I know that I can use a similar argument to prove, for instance, $\mathbb{Z}_4 \trianglelefteq S_4$ by observing that $\mathbb{Z}_4$ is the set of all permutations in $S_4$ which are length-4 cycles. [i.e. $1 \simeq (1234), 2 \simeq (1324)$, ...].

I think I can use this type of reasoning to prove $V_4 \trianglelefteq S_4$, (where $V_4$ is the Klein 4-group) by characterizing $V_4$ as the set of permutations in $S_4$ that are a product of $0$ or $2$ disjoint transpositions. Cycle type and disjointness are both preserved by relabeling the elements. (Is this correct?)

My question is: what is special about $S_4$ in the above? Aren't any permutations "relabelings"? They may be a certain type of relabeling, but it seems that the above argument would still work. Does it?

For example, can I show that $V_4 \trianglelefteq A_4$ by a similar argument to the above? In general, any group is a group of permutations. So if a subgroup of some group $G$ is exactly a certain set of cycle types, is it automatically normal?

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Eli Rose
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  • I think your overall thought process is good, but note that $S_4$ doesn't have any normal subgroups isomorphic to $\Bbb Z_4$. The set of permutations that consist of a single $4$-cycle has cardinality $6$ in fact, and together with the identity, doesn't form a subgroup. – pjs36 Mar 23 '16 at 16:16
  • @pjs36: Oops! Thanks. So obvious it has a non-normal subgroup isomorphic to $\mathbb{Z}_4$, but not a normal one. – Eli Rose Mar 23 '16 at 16:17

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I think you're essentially using the following theorem (that isn't too hard to prove):

Let $G$ by a group, and $H \leq G$ a subgroup of $G$. Then $H$ is normal in $G$ if and only if $H$ is a union of conjugacy classes of $G$.

This pairs incredibly well with symmetric groups, as the conjugacy class of a permutation $\tau \in S_n$ is, as you've pointed out, all permutations of $S_n$ with the same cycle structure as $\tau$. We can rephrase the above theorem then, for symmetric groups, to say that

A subgroup $H$ of the symmetric group $S_n$ is normal in $S_n$ if and only if, whenever it contains a permutation $\tau$, it contains the entire conjugacy class $\operatorname{cl}_{S_n}(\tau)$, that is, all permutations with the same cycle structure.

  • This applies really nicely to the alternating group $A_n$, since it consists of all permutations with an even number of cycles of even length. This isn't a single conjugacy class of elements, of course, but it neatly decomposes into a union of conjugacy classes.
  • It applies nicely to the normal Klein $4$-group $V = \{(1),\, (12)(34),\,(13)(24),\,(14)(23)\}$ of $S_4$.
  • You'll fail to generate a proper normal subgroup of $S_n$ if you try to include all the $4$-cycles, since then you'll start generating all $2$-cycles and you'll have enough permutations to generate all of $S_n$.

Note that normality of $V$ in $S_4$ automatically implies normality in any subgroup of $S_4$, namely $A_4$. You could dig into conjugacy classes of $A_4$ and appeal to the main theorem, if you wanted to though. There's some messiness in trying to use this approach in $A_n$, as some conjugacy classes of $S_n$ split into two conjugacy classes in $A_n$, while others don't (see here for instance). It still works, it's just a bit more work to keep track of conjugacy classes.

The above theorem about normal subgroups being a union of conjugacy classes is probably the most elementary way to prove that the alternating group $A_5$ is simple. Its conjugacy class sizes are $1,\,12,\,12,\,15$, and $20$: There are no ways to add these numbers up (including the $1$, for the identity!) and get a divisor of $60$, hence no nontrivial, proper, normal subgroups.

pjs36
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  • It's possible that I've completely missed the spirit of your question! If so, do say something, as hopefully you can still get the answer you're looking for. – pjs36 Mar 23 '16 at 17:17
  • Nice answer (+1). Minor typo: I assume you meant "prove that the alternating group $A_5$ is simple" –  Mar 23 '16 at 17:22
  • I did indeed mean that, thank you, @Bungo ! – pjs36 Mar 23 '16 at 17:23
  • Thanks for the detailed answer! I wish we covered this stuff about conjugacy classes. So I think that maybe what my intuition is trying to use is "normality in $S_4$ implies normality in any subgroup of $S_4$". I was unaware of that -- is it generally true that if $G \trianglelefteq K$, and a subgroup of $H$, then $H \le K$ implies $G \trianglelefteq H$? – Eli Rose Mar 23 '16 at 17:55
  • @EliRose Yes, that's right! I'm a bit thrown off by the notation, but let's say $N$ is normal in $G$, with $N$ contained in a subgroup $H$ of $G$. Since $N$ is normal in $G$, conjugating things in $N$ by any element of $G$ lands us back in $N$ (that is, $g^{-1}ng \in N$ for all $n \in N,\ g \in G$). Then conjugating things in $N$ by things in $H \leq G$ will certainly leave us in $N$ still, since things in $H$ are also in $G$, and we just checked conjugation by everything in $G$, including those in $H$! – pjs36 Mar 23 '16 at 18:07
  • Of course! That's beautiful and certainly makes sense. Thanks! – Eli Rose Mar 23 '16 at 18:33
  • So, any subgroup that can be described in a way that is invariant under relabeling is normal in $S_n$, hence it is normal in whatever it's a subgroup of (since all groups are subgroups of some $S_n$). – Eli Rose Mar 23 '16 at 18:37
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    @EliRose I needed some time to think about that one, but I think it's a pretty good heuristic, I have to agree. – pjs36 Mar 24 '16 at 03:55
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It is well-known that the 2-cycles generate $S_n$, the 3-cycles generate $A_n$. In general, one can prove this: for $1 \lt k \leq n$, the subgroup generated by all the $k$-cycles equals $S_n$ if $k$ is even, and equals $A_n$ if $k$ is odd.

Nicky Hekster
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