2

A sequel to my earlier question:

I've been doing some more concrete arithmetic with one of these outer automorphisms and it's working out just the way "they" say it will (e.g. $(123)(45)(6)$ goes to $(ABCDEF)$ and $(123456)$ goes to $(A)(BC)(DEF)$, etc. etc.).

My next question: If I'm not mistaken, $S_6$ should be a subgroup of some larger group of permutations, but smaller than the group of all permutations on that larger set---I would think a group of permutations on a set that has $\{1,2,3,4,5,6\}$ as a subset---such that some inner automorphism of that larger group, when restricted to $S_6$, is an outer automorphism of $S_6$. How big is that larger set and what group of permutations on it do we need to look at, and which of its members do we conjugate by?

Community
  • 1
Michael Hardy
  • 1
  • Interesting. I'm wondering if in general given some outer automorphism of a group $G$, could you embed $G$ as a normal subgroup of some group $H$, where conjugation by an element of $H$ gives the outer automorphism? For example outer automorphisms of $A_n$ are conjugation by elements of $S_n$ when $n \neq 2, 3, 6$. – Mikko Korhonen Mar 25 '12 at 21:15
  • I believe that does hold generally. – Michael Hardy Mar 25 '12 at 21:16
  • 2
    Yes. Just take the semidirect product of $S_6$ with $\mathrm{Aut}(S_6)$ :) – Bill Cook Mar 25 '12 at 21:24
  • 1
    For "concrete arithmetic" with groups, you might consider GAP. You can easily compute the outer automorphism of $S_6$ explicitly, as explained by Alexander Hulpke here. In particular, you can get it as $\sigma$ of order 10, mapping $(5,6)$ to $(5,6)^\sigma = (1,2)(3,5)(4,6)$, leaving $(1,2,3,4,5)$ fixed, or as an automorphism of order 2, mapping $(1,2,3,4,5,6)\mapsto (2,6)(3,5,4)$ and $(1,2)\mapsto (1,2)(3,4)(5,6)$, etc. (If these don't agree with what you see, keep in mind that in GAP operations act on the right.) – William DeMeo Mar 25 '12 at 21:54
  • P.S. my previous comment doesn't answer your question, but I think Bill Cook's does. – William DeMeo Mar 25 '12 at 22:05
  • $\mathrm{Aut}(S_6)$ is also connected to the Mathieu group $M_{12}$ on twelve letters; $M_{12}$ acts naturally on a set of 12 letters. – Arturo Magidin Mar 25 '12 at 22:12
  • @BillCook : What little I know of semidirect products is not sufficient for me to say anything confidently without first looking at that infallible (but fallible) source called Wikipedia, which gives one characterization thus: if $N$ and $H$ are subgroups of $G$ and $G=NH$ and $N\cap H = \varnothing$, then $G$ is a semidirect product of $N$ and $H$. This by no means makes it clear that there is such a thing as "the" semidirect product, nor how to "take" it if it doesn't fall into your lap. (I first saw the concept in a geometry course, where it was pointed out that the group of....... – Michael Hardy Mar 25 '12 at 23:31
  • ......isometries of Euclidean space is a semi-direct product of the group of isometries fixing the origin and the group of translations.) – Michael Hardy Mar 25 '12 at 23:32
  • 1
    @MichaelHardy: You want to look at the external semidirect product, which requires a group $N$, and a homomorphism $\varphi\colon H\to \mathrm{Aut}(N)$; then you define $N\rtimes_{\varphi} H$ as the set of pairs $(n,h)\in N\times H$ with product $$(n_1,h_1)(n_2,h_2) = (n_1\varphi(h)(n_2), h_1h_2).$$ With $H=\mathrm{Aut}(N)$, you get a natural action (via $\varphi=\mathrm{id}$), which creates a group $N\rtimes \mathrm{Aut}(N)$ known as the holomorph of $N$. – Arturo Magidin Mar 26 '12 at 00:00
  • @ArturoMagidin Thanks! Couldn't come up with the name "holomorph" :) – Bill Cook Mar 26 '12 at 01:28
  • OK, I'm digesting the answers below. I shall return...... (Or, as someone once said, I'll be back.....) – Michael Hardy Mar 27 '12 at 18:49

2 Answers2

5

As noted by Bill Cook, there is a natural group to look at, which is the holomorph of $S_6$.

This works for any group.

More generally: if $N$ is a group, and $\varphi\colon H\to \mathrm{Aut}(N)$ is a group homomorphism, then this induces an action of $H$ on $N$ by the formula $${}^hn = [\varphi(h)](n),$$ where "${}^hn$" means "the image of $n$ under the action of $h$". The formula makes sense because $\varphi(h)\in\mathrm{Aut}(N)$, so we can evaluate $\varphi(h)$ on $n$.

This action satisfies nice properties:

  • ${}^h(nm) = {}^hn{}^hm$ for all $n,m\in N$, $h\in H$.
  • ${}^k({}^hn) = {}^{kh}n$.

Given such a triple, $(N,H,\varphi)$, we can construct a $G$ that contains subgroups $\mathcal{N}$ and $\mathcal{H}$ isomorphic to $N$ and $H$, respectively, with $\mathcal{N}\triangleleft G$, $G=\langle\mathcal{N},\mathcal{H}\rangle=\mathcal{NH}$, and where for every $\mathfrak{h}\in \mathcal{H}$ and $\mathfrak{n}\in\mathcal{N}$, if $h\in H$ corresponds to $\mathfrak{h}$ and $n$ corresponds to $\mathfrak{n}$, then $\mathfrak{hnh}^{-1}$ corresponds to ${}^hn$. This is the semidirect product $N\rtimes_{\varphi}H$.

The underlying set of $N\rtimes_{\varphi}H$ is $N\times H$. The multiplication rule is $$(n_1,h_1)\cdot (n_2,h_2) = \Bigl(n_1{}^hn_2,h_1h_2\Bigr).$$ I'll leave it to you to verify this group satisfies the properties given above, with

  • $\mathcal{N} = \{ (n,1)\mid n\in N\}$;
  • $\mathcal{H} = \{(1,h)\mid h\in H\}$.

Note that, indeed, we have: $$(1,h)(n,1)(1,h)^{-1} = (1{}^hn,h)(1,h^{-1}) = ({}^hn,h)(1,h^{-1}) = ({}^hn{}^h1,hh^{-1}) = ({}^hn,1).$$

Conversely, if $G$ is a group, $N\triangleleft G$, $H\leq G$ are subgroups with $NH=\langle N,H\rangle = G$ and $N\cap H=\{1\}$, then for each $h\in H$ we have an automorphism of $N$ given by $n\mapsto hnh^{-1}={}^hn$; this yields a group homomorphism $\varphi\colon H\to \mathrm{Aut}(N)$, and it is easy to verify that the group $N\rtimes_{\varphi}H$ constructed as above is isomorphic to $G$.

Now, by taking $\varphi\colon\mathrm{Aut}(G)\to\mathrm{Aut}(G)$ being the identity, we can always construct the group $G\rtimes_{\mathrm{id}} \mathrm{Aut}(G)$, which is the aforementioned holomorph of $G$.

Community
  • 1
Arturo Magidin
  • 398,050
5

As others have pointed out, you can embed any group in its holomorph. But for a group $G$ with trivial centre (like $S_6$), the normal subgroup ${\rm Inn}(G)$ of ${\rm Aut}(G)$ is naturally isomorphic to $G$, so you can embed $G$ directly into ${\rm Aut}(G)$, which is of course a smaller group than the holomorph.

In fact ${\rm Aut}(S_6)$ embeds into $S_{12}$ as follows. Let $\tau$ be an outer automorphism of ${\rm Aut}(S_6)$ with $\tau^2=1$. For an element $g \in S_6$ acting on the set $\{1,2,3,4,5,6\}$, let $\tau'(g)$ denote $\tau(g)$, but acting on the set $\{7,8,9,10,11,12\}$. So, for example, if $g=(1,2)$ and $\tau(g) = (1,2)(3,4)(5,6)$, then $\tau'(g) = (7,8)(9,10)(11,12)$.

Then we can embed $S_6$ into $S_{12}$ by $g \to g\ \tau'(g)$. Then $t := (1,7)(2,8)(3,9)(4,10)(5,11)(6,12)$ induces $\tau$ by conjugation on the image of this embedding, so $\tau$ together with this image generates ${\rm Aut}(S_6) \le S_{12}$.

This image is a subgroup of the Mathieu group $M_{12}$ and can be used as part of one of the many constructions of $M_{12}$.

Derek Holt
  • 90,008