I am working on a project in my group theory class to find an outer automorphism of $S_6$, which has already been addressed at length on this site and others. I have a prescription for how to go about finding this guy, but I have a larger conceptual question - what does an outer automorphism really look like? Is there an intuitive way to understand the difference between an inner and an outer automorphism? Inner automorphisms have always seemed easier for me to understand since we have an explicit representation $ghg^{-1}$ for members of the group. I can also understand this representation in terms of the Rubik's cube - rotate an edge, rotate a perpendicular edge, and the rotate the other edge back (is not the same as just rotating the perpendicular edge). What does an "outer automorphism" look like?

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    $\begingroup$ If you want a huge class of outer automorphisms, just look at abelian groups with nontrivial automorphisms, such as vector spaces, especially those of dimension greater than $1$. $\endgroup$ – Lubin Oct 20 '12 at 21:55
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    $\begingroup$ I was just about to suggest Lubin's example - the very simplest 'outer' automorphism may be the simple $x\to -x$ on $(\mathbb{Z}, +)$. $\endgroup$ – Steven Stadnicki Oct 23 '12 at 17:30
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    $\begingroup$ @StevenStadnicki this should be the example that introductory textbooks provide. It is a lot more accessible than analyzing $S_6$ $\endgroup$ – tacos_tacos_tacos Apr 12 '14 at 6:15

Since you're working on $S_6$ for class, I won't answer that one for you. But I have another easy example of an outer automorphism.

The Quaternion group $Q_8$ can be represented as $\{1, -1, i, -i, j, -j, k, -k\}$, where $ij=k$, $jk=i$, $ki=j$, and $ji=-k$, $kj=-i$, $ik=-j$. The $-1$ element acts pretty obviously on the rest, e.g. $(-1)j=-j$. I think conceptually this is an easy presentation to work with.

So, $Q_8$ has $4$ proper nontrivial subgroups: $Z=\{1,-1\}$ (which is characteristic, and equal to both the center and commutator subgroups), $I=\{1,-1,i,-i\}$, $J=\{1,-1,j,-j\}$, and $K=\{1,-1,k,-k\}$.

All of these subgroups are normal, so there is no element in $x \in Q_8$ for which $I^x=J$, for example. Perhaps an inner automorphism will permute elements nontrivially within those three subgroups, but it won't move the subgroups themselves.

However, it's pretty clear just from looking at $I$ $J$ and $K$ that there's no real difference between these subgroups. If we changed $i$'s name to $j$, $j$'s name to $k$, and $k$'s name to $i$, this wouldn't mess up the group in any way - so, we can expect that a mapping $\sigma:i\mapsto j, j\mapsto k, k\mapsto i$ defines an automorphism of $Q_8$. In other words, this outer automorphism permutes the set of subgroups $\{I,J,K\}$. As a matter of fact, the outer automorphism group of $Q_8$ turns out to be $S_3$, and acts by permuting this set of three subgroups.

EDIT: I wanted to add another thing, a heuristic I use when thinking about automorphisms in general. I find it helpful, for some reason, to think about contructing an isomorphism from $G$ to $G$ (as if $G$ were some different group I was trying to prove isomorphic to $G$). To me this is very intuitive especially when considering outer automorphisms, as inner automorphisms do have that concrete definition using group elements, but outer automorphisms don't. This helped me to better understand what "characteristic" subgroups are - for example, any automorphism (outer or not) of $Q_8$ had better map $-1$ to $-1$, because it's the only element that acts like that. The same holds for the center of any group, the commutator subgroup of any group, normal (and hence unique) Sylow subgroups, any term of the derived, upper central, lower central, fitting, or frattini series, and so on. Anything that is uniquely defined using only properties of the group must be fixed by any automorphism, because if $G$ is isomorphic to $G$, the isomorphism between them must send those subgroups to the same place. Note that, as the comment points out, this heuristic requires some additional thought when the group is not finite.

  • $\begingroup$ With respect to the edit, you have to be careful constructing isomorphisms between infinite groups. That is, with a finite group you just have to find a surjective homomorphism from $G$ to $G$ (so if you are working with presentations, words on the generators which respects the presentation and generate the group). However, with infinite groups such a map is not necessarily injective! If there exists such a map which is non-injective then $G$ is called non-Hopfian (else, they're called Hopfian). $\endgroup$ – user1729 Oct 22 '12 at 12:03
  • $\begingroup$ @user1729 if $G$ is finitely generated then is it "Hopfian?" If $G$ is countably infinite is it Hopfian? $\endgroup$ – tacos_tacos_tacos Nov 13 '12 at 15:07
  • $\begingroup$ @jshin47: No. See this thread. The "obvious" (but infinitely presented) example is when you take infinitely many copies of a group and cross product them together. Killing the left-most group results in an isomorphic group. A more subtle example is the group with presentation $\langle a, t; t^{-1}a^2t=a^3\rangle$. Here, the map $a\mapsto a^2$, $t\mapsto t$ has non-trivial kernel but is onto. $\endgroup$ – user1729 Nov 13 '12 at 21:50
  • $\begingroup$ I wonder if the converse is true, regarding "unique" elements or subgroups within a group: If a group contains two or more subgroups that are isomorphic to each other, does this guarantee the existence of automorphisms (inner or outer) that maps these subgroups to each other? $\endgroup$ – Nicolas Miari Aug 22 '16 at 5:24

In some sense, you've basically figured out the key difference between the two. The inner automorphisms are the easy automorphisms to construct. Conjugating by an element is a way to produce an automorphism in any group. So you go ahead and do that with your group $G$ to produce a bunch of automorphisms.

Now you're left with the question: are there any more? Of course, if you find an automorphism $\phi$ that is not inner, then $\phi \circ \text{Inn}(g)$ will be an automorphism for any $g \in G$. So we may as well try to classify the quotient $\text{Out}(G) := \text{Aut}(G) / \text{Inn}(G)$. But at this point, there is no general approach that will work for all $G$. You just have to get your hands dirty and start studying lots of example.

As you no doubt know, what makes $S_6$ so special is that it does have a non-trivial automorphism (essentially only one because $\text{Out}(G) \cong \mathbb{Z} / 2\mathbb{Z}$), yet no other $S_n$ has an outer automorphism. All I've been able to do is marvel at how the universe allows an outer automorphism when $n=6$ but never for any other value of $n$. But I don't have a better explanation besides an outer automorphism has been constructed for $S_6$ and it's also been proven that one cannot be constructed for any other $S_n$.

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    $\begingroup$ The possibility of an outer automorphism when $n=6$ arises from a numerical coincidence. The order of the centralizer of a transposition in $S_n$ is $2(n-1)!$. For all $n \ne 6$, you can show that all involutions with centralizer of that order are transpositions. Hence all automorphisms map transpositions to transpositions, from which it follows that they must all be inner. But when $n=6$ involutions in the class $(1,2)(3,4)(5,6)$ happen to have centralizers with the same order, and it turns out that there is indeed an outer automorphism mapping $(1,2)$ to $(1,2)(3,4)(5,6)$. $\endgroup$ – Derek Holt Oct 20 '12 at 21:45
  • $\begingroup$ @DerekHolt: Thanks for the helpful explanation of why $n=6$ is different! $\endgroup$ – Michael Joyce Oct 21 '12 at 12:14

Another rather accessible example of an outer automorphism can be found on matrix groups: Look at the inverse transpose map $$\operatorname{GL}(n,F) \to \operatorname {GL}(n,F), \quad A \mapsto (A^{-1})^\top.$$ It is an automorphism, and except from all cases with $n=1$ and $\operatorname{GL}(2,\mathbb F_2)$, it is is outer.


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