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Let $K$ be a field, then $\operatorname{GL}(n,K)$ consists of the $n\times n$ invertible matrices, $\operatorname{SL}(n,K)$ consists of the $n\times n-$matrices with determinant $1$, and $\operatorname{PGL}(n,K) = \operatorname{GL}(n,K)/\{rI_n: r\in K^\times\}$, $\operatorname{PSL}(n,K) = \operatorname{SL}(n,K)/\{rI_n: r^n=1\}$.

Now I'm interested in the automorphism groups $\operatorname{Aut}(\operatorname{SL}(n,K))$, $\operatorname{Aut}(\operatorname{PGL}(n,K))$, and $\operatorname{Aut}(\operatorname{PSL}(n,K))$. I've searched for small $n$ and $|K|$ using GAP and found that they are the same. So are these three groups really isomorphic?

Edit 1: I realized that $\operatorname{SL}(n,K)$, $\operatorname{PGL}(n,K)$, and $\operatorname{PSL}(n,K)$ are in many cases isomorphic by themselves. For $K$ finite, this happens if and only if $\gcd(n,|K|-1)=1$ (so for $n=3$, the first nontrivial cases are $|K|=4$ and $|K|=7$).

Edit 2: I think my conjecture is very likely to fail if $K$ behaves not so well. So please assume that $K$ is finite if necessary.

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  • $\begingroup$ If $\gcd(n,|K|-1)=1$ then $GL(n,K)\cong SL(n,K)\times K^*$ (as abstract group, not algebraic group) so $Aut(GL(n,K))$ is larger than $Aut(SL(n,K))$ $\endgroup$
    – reuns
    Apr 4, 2022 at 14:22

1 Answer 1

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It is true in general that $\text{Aut}(SL_n(K)) \cong \text{Aut}(PSL_n(K)) \cong \text{Aut}(PGL_n(K))$ are naturally isomorphic for any (commutative) field $K$ and any dimension $n$. However, $GL_n(K)$ usually has additional central automorphisms owing to the fact that is has plenty of nontrivial maps from its abelianization to its center. The extra automorphism can be identified with a subset of $\text{End}(K^\times)$, but I don't believe it is understood well for general $K$.

A reference is Dieudonné, La géométrie des groupes classiques (see the automorphism group of the general linear group GL(n,F) )


One way to track down outer automorphisms of a group $G$ is to study its embeddings inside a larger group $\widetilde G$, then restrict inner automorphisms from $N_{\widetilde G}(G)$ to $G$. One obtains a sequence:

$$ 1 \to Z(G) \cap C_{\widetilde G}(G) \to C_{\widetilde G}(G) \to N_{\widetilde G}(G)/G \to \text{Out}(G) $$

(Any outer automorphism arises this way because of $\widetilde G = G \rtimes \text{Aut}(G)$.)

In particular one can hope to find $\widetilde G$ such that the last map is onto, which is exactly what is done for $\text{Aut}(PSL_n(K))$ in the original work of Schreier-van der Waerden (1928). As mentioned in Dieudonné, there is an error in logic when $n=2$ and characteristic is $\neq 2$, but later work showed the result was still correct.

S-W identify the automorphism group with the projective correlation-collineation group, which is isomorphic to an extension of the projective semilinear group, $\mathit{P\Gamma L}_n(K)\rtimes C_2$, with the transposition acting by the transpose-inverse map (similarly to the outer linear group). However, for $n = 2$ the extra factor is not needed since transpose-inverse satisifies:

$$X^{-T} = (\text{det}\, X)^{-1}CXC^{-1}$$

for $C = e_{21} - e_{12}$, and therefore is an inner automorphism of $SL, PSL, PGL$.

Theorem 1 (Schreier-van der Waerden). For a field $K$, the automorphism group of $PSL_n(K)$ is naturally isomorphic to $$ \text{Aut}(PSL_n(K)) \cong \begin{cases} \mathit{P\Gamma L}_n(K) & n = 2\\ \mathit{P\Gamma L}_n(K) \rtimes C_2 & n \geq 3 \end{cases} $$ via the map induced by the inclusion $PSL_n(K) \to \mathit{P\Gamma L}_n(K)$.

The disappearance of the $C_2$ is actually somewhat natural from the point of view of projective geometry. This is because the transpose-inverse map represents the "standard duality" swapping points with hyperplanes (cf. correlation), but for $n=2$ this is projective line, so the duality is trivial.

The conjugate transpose can also be viewed as the graph automorphism coming from the root system used to construct $PSL_n$ as a Chevalley group. This is the reflective symmetry of the Dynkin Diagram for $A_n$ (which becomes trivial for $n = 2$) and inducing $g \mapsto -g^T$ in the Lie algebra.


In Dieudonné (1951), the results above are generalized to $SL_nK, PGL_nK$ and $GL_nK$ (and orthogonal and symplectic groups) over a division rings, with some restrictions on the dimension and characteristic. However, he mentions in the introduction that the commutative case for all three follow easily from Schreier-van der Waerden's work. I haven't read S-W in detail (it is in German) so I can't say if the proof applies verbatim to $PGL_nK$ for example. However, I am able to deduce the results for these other groups from that of $PSL_nK$ without much effort.

Recall we have an exact square:

$$ \require{AMScd} \begin{CD} \mu_n(K) @> >> K^\times @>{(\cdot)^n}>> (K^\times)^n\\ @VVV @VVV @VVV\\ SL_n(K) @>>> GL_n(K) @>{\text{det}}>> K^\times\\ @VVV @VVV @VVV\\ PSL_n(K) @>>> PGL_n(K) @>{ }>> K^\times/(K^\times)^n \end{CD} $$

where $\mu_n(K) := \{\xi \in K^\times: \xi^n = 1\}$. Also, the determinant map (and its projective version) are split by e.g. $\lambda \mapsto \text{diag}(\lambda, I_{n-1})$, giving semi-direct product decompositions.

This square is natural since the subgroups are centers and (almost always) derived subgroups. We have:

Proposition. The natural inclusion $$Z(GL_n(K)) \leq C_{GL_n(K)}(SL_n(K))$$ is equality, equal to the group of scalar matrices.

(Note: This follows by the same proof at groupprops, except that there it neglects to notice that commuting with all elementary transvections actually implies the matrix is scalar, not merely diagonal.)

Proposition. For the chain of natural inclusions $$[SL_n(K), SL_n(K)] \leq [GL_n(K), GL_n(K)] \leq SL_n(K)$$ we have:

  • Equality on the right, except in the case of $SL(2,2)$
  • Equality on the left, except in the case of $SL(2,3)$

In particular $SL_n(K)$ is a characteristic subgroup. (The same applies to the projective groups.)

When both equalities hold, $SL_n(K)$ is a perfect group, and when the right equality holds $SL_n(K)$ is the derived group of $GL_n(K)$. The sole exception to the latter is $SL_2(\mathbb F_2) = GL_2(\mathbb F_2) \cong S_3$, whose derived group is $A_3 \cong C_3$. On the other hand, it is the first inclusion which is not an equality for $SL_2(\mathbb F_3) \cong Q_8 \rtimes C_3$. These exceptions won't bother us again since in the first case all groups are the same, and in the latter case it so happens that $\text{Hom}(G_{ab}, ZG) \cong \text{Hom}(C_3, C_2) = 1$.


Now to relate the result for $PSL_nK$ to the other groups, note that because $SL_n(K)$ (and scalar matrices) are characteristic subgroups, we have natural restriction/quotient maps:

$$ \require{AMScd} \begin{CD} \text{Aut}(GL_n(K)) @> >> \text{Aut}(PGL_n(K))\\ @VVV @VVV\\ \text{Aut}(SL_n(K)) @>>> \text{Aut}(PSL_n(K)) \\ \end{CD} $$

From Theorem 1, the maps in this diagram ending at $PSL_n$ are all surjective. For example, $GL_n$ and $SL_n$ embed similarly into the nonprojective semilinear group $\Gamma L_n(K)$, and taking inner automorphisms gives us again $P\Gamma L_n(K)$. So, we only need to see injectivity.

Recall that for a group $G$, the central automorphism group $\text{Aut}_c(G)$ is defined to be the kernel of the natural map $\text{Aut}(G) \to \text{Aut}(G/ZG)$ which is by definition the kernel of our horizontal maps above. It is easy to show that a central endomorphism is of the form $[g \mapsto \chi(g)g]$ for $\chi \in \text{Hom}(G_{ab}, ZG)$. When $SL_n(K)$ is perfect, this group vanishes. This also vanishes in the two exceptional cases, for example since $SL(2,3)_{ab} \cong C_3$ but $Z(SL(2,3)) \cong C_2$. So the bottom map is injective.

As for the vertical maps, we can show that in this case, the kernels happen to also be the respective central automorphism groups.

Proposition. Let $G$ be a group such that $G \cong H \rtimes T$, where $H$ is a characteristic subgroup such that $C_G(H) = Z(G)$. Then the sequence $$1 \to \text{Aut}_c(G) \to \text{Aut}(G) \longrightarrow \text{Aut}(H)$$ is exact.

Proof. Given $\Phi \in \text{Aut}_H(G)$ extending $\text{id}_H$, we set $\chi(x) := x^{-1}\Phi(x)$ which satisfies $\chi(h.t) = \chi(t)$. Then we compute:

$$ \begin{aligned} \Phi(t \cdot h) &= \Phi(tht^{-1}.t)\\ &= tht^{-1}\Phi(t)\\ &= t\cdot h\cdot \chi(t)\\ \Phi(t)\cdot\Phi(h) &= \Phi(t)h\\ &= t\cdot \chi(t) \cdot h \end{aligned} $$ Hence $[h, \chi(t)] = 1$, or $\chi(t) \in C_G(H)$. From $C_G(H) \subset Z(G)$, it then follows that this is a central homomorphism. $\square$

Since $PGL_n(K)$ is centerless, its central automorphisms vanish too and the right vertical map is an injection.

Corollary. For any field $K$, we have natural isomorphisms $$\text{Aut}(PGL_nK) \cong \text{Aut}(SL_nK) \cong \text{Aut}(PSL_nK)$$ and $$\text{Aut}(GL_nK) \cong \text{Aut}_c(GL_nK)\rtimes\text{Aut}(PSL_nK)$$

The inner/outer automorphism structure splits up differently depending on $GL$ vs $SL$ via:

$$ \begin{aligned} \mathit{P\Gamma L}_nK &\cong PGL_nK \rtimes \text{Aut}(K)\\ &= PSL_nK \rtimes (K^\times/(K^\times)^n) \rtimes \text{Aut}(K) \end{aligned} $$

Note that the factor of $K^\times/(K^\times)^n$ are the diagonal automorphisms in the classification in Steinberg (cf. Automorphisms of Linear Groups (1960))


We have addressed the main question, no we look at $\text{Aut}_c(GL_n(K))$ only for a finite field.

It's worth noting that although there is a correspondence (called the Adney-Yen mapping), $$\text{Aut}_c(G) \hookrightarrow \text{Hom}(G, ZG)$$ defined by $\phi \mapsto [g \mapsto g^{-1}\phi(g)]$, this is not in general a bijection. Instead the right hand side is in bijection with central endomorphisms $\text{End}_c(G)$, which is in general larger than $\text{Aut}_c(G)$. We can characterize the image:

Proposition. Let $\chi \in \text{Hom}(G,ZG)$ and $\phi \in \text{End}_c(G)$ be such that $\phi(g)g^{-1} = \chi(g)$. Then

  1. $\phi$ is injective if and only if $\chi(g) = g^{-1}$ implies $g = 1$
  2. $\phi$ is surjective if and only if $\chi(g)^{-1} = \chi(z)z$ for some $z \in ZG$

Any two such $z, z'$ in (2.) satisfy $\phi(z) = \phi(z')$, so that for automorphisms this element is unique, and in this way we obtain the element $\overline{\chi} \in \text{Hom}(G, ZG)$ corresponding to $\phi^{-1}$.

Consider a finite field $|K| = q + 1$, then $K^\times \cong C_q$ and $\text{Hom}(G, ZG) \cong \text{End}(C_q) \cong C_q$, given by $m\in C_q \longleftrightarrow [A \mapsto (\text{det} A)^mA]$ with composition on the right corresponding to the operation $m_1 \circ m_2 = m_1 + m_2 + nm_1m_2$ in $C_q$ (the $n$ comes from the determinant). It suffices to check injectivity, which can be checked on scalar matrices and is equivalent to $\lambda^{nm+1} = 1 \iff \lambda = 1$, or $\text{gcd}(nm + 1, q) = 1$.

Let $a = \text{gcd}(n, q)$ and factor $q = \hat ab$ where $\hat a$ is the maximal divisor with $\text{rad}(\hat a) = \text{rad}(a)$ (and $b$ is the maximal divisor such that $\text{gcd}(a, b) = 1$). By the remainder theorem $\mathbb Z/q\mathbb Z = \mathbb Z/\hat a\mathbb Z \times \mathbb Z/b\mathbb Z$, such that $nm+1 = (n_am_a + 1, n_bm_b + 1)$. All elements of the form $n_am_a + 1$ in $\mathbb Z/{\hat a}\mathbb Z$ are units in the coset $1 + a\mathbb Z/{\hat a}\mathbb Z$, whereas all elements of $\mathbb Z/b\mathbb Z$ (in particular the units) are of the form $n_bm_b + 1$ by invertibility of $n$ mod $b$.

Thus $ \text{Aut}_c(GL_n(q+1)) \cong (1 + a\mathbb Z/\hat a\mathbb Z) \times (\mathbb Z/b\mathbb Z)^\times\\ $ and the factor on the left is also the kernel of $(\mathbb Z/(\hat a/a)\mathbb Z)^\times \to (\mathbb Z/\text{rad}(\hat{a}/a)\mathbb Z)^\times$.

In other words, we get the full unit groups coming from the primes dividing $q$ but not $n$, and for the power two, but only the $p$-torsion part for odd primes $p$ diving both $q$ and $n$.

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  • $\begingroup$ Very detailed answer indeed. I like it! Thanks a lot for your elaboration. $\endgroup$ Apr 7, 2022 at 3:24
  • $\begingroup$ I have just one question: I have heard that $\operatorname{Aut}(\operatorname{PGL}(2,p)) = \operatorname{PGL}(2,p)$ for primes $p$, so is the case $n=2$ special? it seems that the extra $C_2$ factor does not exist in this case? $\endgroup$ Apr 7, 2022 at 4:06
  • $\begingroup$ Also for the $\operatorname{GL}$ case, GAP gives $|\operatorname{Aut}(\operatorname{GL}(3,4))| / |\operatorname{PGL}(3,4)| = 12$, so could you please check the formula for $\operatorname{GL}$ :) $\endgroup$ Apr 7, 2022 at 4:58
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ Apr 7, 2022 at 10:24
  • $\begingroup$ Apologies - the factor in GL corresponding to radial automorphisms is simply not correct, and also the transpose map is inner for dimension 2. I will edit the answer when i have a chance but it may be a couple days $\endgroup$
    – Ben
    Apr 8, 2022 at 11:55

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