While reading a paper, I came across the following peculiar condition:

Let $1 \rightarrow H \rightarrow G \rightarrow G/H \rightarrow 1$ be a short exact sequence, and let $H$ be abelian. We require that any automorphism, $\sigma$, of $G$ that preserves $H$ pointwise and such that $\sigma(g)H=gH$ (preserves cosets pointwise) is trivial.

This doesn't give me a good clue as to what we're talking about. For example: is it true for any semi-direct product of two groups of coprime order (the first one being abelian, as required)? How about for a semi-direct product of cyclic groups of coprime order? Prime coprime orders? What is this condition, and what are some canonical examples of it?

I apologize if this is simple, but my finite group theory is pretty shoddy.

  • $\begingroup$ $S_3$ is a semidirect product of two cyclic groups of relatively prime orders, but conjugation by $(123)$ doesn't satisfy this condition. This condition implies (but is not equivalent to) the first cohomology being zero. $\endgroup$ – Steve D May 30 '10 at 21:41

Short answer: a typical example is G=SL(2,5), H = Z(G) = Z/2Z. If G/H and H are coprime and satisfy the condition, the G = G/H × H is quite dull.

I'll assume you find this interesting, and want to read about it:

Let G be a group (finite is good), H be an abelian normal subgroup of G, and Q be the quotient group.

If σ is an automorphism of a group G such that σ(H) = H, then σ induces automorphisms on H and G/H=Q.

If σ(h)=h for all h in H, then σ(H) = H, so we are interested in those σ that are "invisible" as automorphisms both of H and of Q.

The paper's condition is that no automorphism is invisible, which should seem a reasonable crutch if you want to talk about automorphisms of G in terms of those of H and Q (here it helps if H is also characteristic).

What do invisible automorphisms look like?

Well every element of G can be written as a product q*h for some q in Q and h in H. (q1*h1)*(q2*h2) = (q1*q2)*(h1^q2 * h2 * ζ(q1,q2)) where ζ:Q×Q→H is a (set-theoretic) function called a 2-cocycle. If you are only interested in semidirect products, then ζ(q1,q2) = 1H can be ignored.

What does σ do to q*h? Well it takes products to products, and h to h, but it only takes q to another element of the same coset, q*δ(q) where δ:Q→H is another set-theoretic function. Hence σ(q*h) = q*h*δ(q).

A good example to keep in mind here are the extra-special groups, like the dihedral group of order 8. They have lots of invisible automorphisms. For instance (x,y)→(x,yz) where x,y are the main generators and z=[x,y] generates the center.

Now clearly not all δ can work, since surely δ(q1*q2) is related somehow to δ(q1) and δ(q2). Indeed σ(q1*q2) = (q1*q2)*δ(q1*q2), but it is also equal to σ(q1)*σ(q2) = (q1*δ(q1))*(q2*δ(q2)) = (q1*q2)*( δ(q1)^q2 * δ(q1) * ζ(q1,q2) ).

Ignoring ζ for a moment, one gets the equation δ(q1*q2) = δ(q1)^q2 * δ(q2), which expresses the fact that δ:Q→H is a derivation of the Q-module H. It is not too hard to see that the implications are reversible, and derivations help to define automorphisms fixing H and Q.

Not ignoring ζ does not change things very much, as instead of the subgroup of derivations inside the abelian group of all functions from Q to H, you just take a coset of this subgroup determined by ζ.

At any rate, so in your semi-direct products the condition is that there are no non-identity derivations from Q to H; the only derivation should have δ(q)=1H for all q in Q.

Now of course derivations can exist even when Q and H are coprime: Take G to be the non-abelian group of order 6, H to be its subgroup of order 3. Then δ:Q→H takes the non-identity element of Q to any one of the three elements of H, giving three derivations δ. Checking the automorphism group of G, it is easy to see that every automorphism must take H to H, and must act as the identity on Q, since Aut(Q) = 1. Of the six automorphisms, three act as inversion on H, so are not inivisible, but three must be invisible, one for each δ.

In particular, being cyclic of prime and coprime order is not sufficient. Your coprime condition does, however, severely limit the variety of invisible homomorphisms that are available: they must all be conjugations by elements of H.

An automorphism of G induced by conjugation by an element of H must act trivially on H, since H is abelian. It must act trivially on Q, since hH = 1Q in Q. Hence every automorphism induced by an element of H is invisible.

If G is to have no invisible automorphisms, then H must be central, since q^h = h^-1 * q * h = q* (h^q)^-1 * h is only the identity on G if h^q = h for all q.

In other words, the paper's condition implies H is central, so you are looking for a group Q with a trivial module H whose first cohomology vanishes, but whose second does not. I think this is reasonably rare. Probably a very standard example is a perfect group Q that is not superperfect, and H to be its Schur multiplier.

For instance, take G=SL(2,5), Q = Alt(5), H = Z/2Z.

  • $\begingroup$ Thank you very much! That was very clear, and very helpful! $\endgroup$ – Makhalan Duff May 30 '10 at 22:10

The condition implies that every stabilizing automorphism of $G$ is trivial. Now the first cohomology group $\text{H}^1(G/H, H)$ is a quotient of the group of stabilizing automorphisms (denoted by $\text{Stab}(G/H,H)$) by the inner automorphisms stabilizing the extensions. $$\text{H}^1(G/H, H)\cong \text{Stab}(G/H,H)/(\text{Inn}(G)\cap \text{Stab}(G/H,H))$$

By the theorem of Zassenhaus group extensions of coprime groups are semi-direct products. Furthermore it can be shown that in this case the cohomology in degree greater than zero is trivial. Thus every stabilizing automorphism is an inner automorphism. A little computation shows that the inner automorphism is conjugation by an element of $H$.


Such automorphisms appear naturally when one tries to analyse the group of automorphisms of $G$ preserving $H$ ($H$ may for instance be a characteristic subgroup so that it is preserved by all automorphisms). Such an automorphism will give one automorphism of $H$ and one on $G/H$ giving a group homomorphism to the product of these two automorphism groups. $\sigma$ lies in the kernel of that map precisely when it fulfils the condition you give. We have a subgroup of that kernel consisting of the conjugations by elements of $H$.

This works for any normal $H$ but in the case when $H$ is abelian it is a classic fact that the group of such $\sigma$ modulo those given by conjugation by elements of $H$ is isomorphic to the cohomology group $H^1(G/H,H)$. In particular if $G/H$ and $H$ have coprime order then all cohomology groups $H^i(G/H,H)$ for $i>0$ are zero and in particular all of your $\sigma$'s are given by conjugations of elements of $H$.

  • $\begingroup$ It is interesting that the set of "invisible" automorphisms forms an abelian subgroup of Aut(G). To be more precise: Let H be any subgroup of G and let A be a group of automorphisms of G fixing each element of H and each right coset of H in G. Then [G,A] is contained in H, so [G,A,A] = 1. Obviously then, [A,G,A] = 1. By the three-subgroups lemma, [A,A,G] = 1. Thus means that the derived subgroup A' = [A,A] acts trivially on G. Since we are working with automorphisms, this says that A' = 1, so A is abelian. There is much more about this sort of thing in Section 4C of my group theory book. $\endgroup$ – Marty Isaacs Jan 31 '12 at 19:33

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