automorphism group of automorphism group of simple group G

Art of Problem Solving

AoPS Online

College Math Topics in undergraduate and graduate studies

Topics in undergraduate and graduate studies

3 M G

BBookmark VNew Topic kLocked

College Math Topics in undergraduate and graduate studies

Topics in undergraduate and graduate studies

3 M G

BBookmark VNew Topic kLocked

w

1 user

TOPICS

superior algebra

B

No tags match your search

M

superior algebra

B

No tags match your search

M

automorphism group of automorphism group of simple group G

G H J

Reveal topic tags

superior algebra

superior algebra unsolved

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

80 posts

#1 h Jan 22, 2015, 9:38 PM • 2 Y

Y by Adventure10, Mango247

$G$ is a non-Abelian finite simple group(its only non-trival normal sub-group is itself).
Proof: $Aut (Aut G) = Inn (Aut G)$

This post has been edited 1 time. Last edited by chenyuandong, Aug 2, 2015, 9:39 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

90 posts

#2 h Jul 16, 2015, 5:24 PM • 1 Y

Y by Adventure10

Because $G$ is simple $\rightarrow Aut(Aut(G))$ is simple $(1)$ .
Now $Inn(Aut(G))$ is a non-trivial normal sub-group of $Aut (Aut(G))$ $(2)$ .
From $(1)$ and $(2)$ we obtain the statements.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

12096 posts

jmerry

#3 h Jul 16, 2015, 9:15 PM • 1 Y

Y by Adventure10

As stated, it's false. Consider, for example, $G$ cyclic of order $13$ . $\mathrm{Aut}(G)$ is cyclic of order $12$ , and $\mathrm{Aut}(\mathrm{Aut}(G))$ is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^2$ , but of course the abelian group $\mathrm{Aut}(G)$ has no nontrivial inner automorphisms.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

227 posts

totem

#4 h Jul 18, 2015, 12:19 PM • 2 Y

Y by Adventure10, Mango247

jmerry wrote:

As stated, it's false. Consider, for example, $G$ cyclic of order $13$ . $\mathrm{Aut}(G)$ is cyclic of order $12$ , and $\mathrm{Aut}(\mathrm{Aut}(G))$ is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^2$ , but of course the abelian group $\mathrm{Aut}(G)$ has no nontrivial inner automorphisms.

Let's consider the group $\mathbb{Z}_{13}$ and try to go through that:

Quote:

$\mathrm{Aut}(G)$ is cyclic of order $12$ ,

That's an interesting phenomenon: If the order of the group is prime $p$ , then there are $p-1$ generators for the group and since these are single-member generators, the mapping $g_1^k\to g_2^k$ (actually $k g_1\to k g_2$ for additive groups) leads to a bijective homomorphism hence $\big|\operatorname{aut} \mathbb{Z}_p\big|=p-1$ .

Quote:

but of course the abelian group $\mathrm{Aut}(G)$ has no nontrivial inner automorphisms.

I assume that's because if the group is abelian then obviously $axa^{-1}=x$ so the only inner automorphism is the identity automorphisms $1_G(x)=x$ .

Quote:

and $\mathrm{Aut}(\mathrm{Aut}(G))$ is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^2$ ,

Not sure about that one. May I ask what $(\mathbb{Z}/2\mathbb{Z})^2$ means? I suspect it's $\mathbb{Z}_2\times \mathbb{Z}_2$ but it's not what I get: First note $\big|\operatorname{aut}\mathbb{Z}_{13}\big|=12$ . So this is a simple abelian group of order 12 and we know a simple abelian group of order 12 is isomorphic to the product group $\mathbb{Z}_{2^4}\times \mathbb{Z}_3$ and it's easy to show:
$\big|\operatorname{aut}\mathbb{Z}_{2^2}\times \mathbb{Z}_3\big|=4$
however that means $\operatorname{aut}\operatorname{aut} \mathbb{Z}_{13}\cong \mathbb{Z}_4$ .

Now, assuming I get this far, how do I actually find the auto-automorphisms? As usual I guess I can start with a brute-force, exhaustive search.

Maybe that would be an interesting project to work on: find $\operatorname{aut}\operatorname{aut} \mathbb{Z}_n^*$ for the first 500 groups and letting $\operatorname{aut}^k \mathbb{Z}_n^*$ represent the k-folded automorphism group, what is smallest $k$ such that $\operatorname{aut}^k\mathbb{Z}_n^*=1_G$ ?

This post has been edited 14 times. Last edited by totem, Jul 18, 2015, 5:06 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

12096 posts

jmerry

#5 h Jul 18, 2015, 7:31 PM • 2 Y

Y by Adventure10, Mango247

totem wrote:

May I ask what $(\mathbb{Z}/2\mathbb{Z})^2$ means?

I prefer $\mathbb{Z}/n\mathbb{Z}$ or $\mathbb{Z}/n$ (spoken "Z mod n") over $\mathbb{Z}_n$ as notation; the latter is potentially ambiguous due to the $p$ -adic integers.

totem wrote:

...however that means $\operatorname{aut}\operatorname{aut} \mathbb{Z}_{13}\cong \mathbb{Z}_4$ .

Just because a group has order $4$ doesn't mean it's cyclic. The automorphisms of that product $\mathbb{Z}/4\times \mathbb{Z}/3$ look like the product of the two automorphism groups.

totem wrote:

...what is smallest $k$ such that $\operatorname{aut}^k\mathbb{Z}_n^*=1_G$ ?

Not going to work. For example, if you start with $n=84$ , the next step is the product of a two-element group and the simple group $GL_3(\mathbb{F}_2)$ of order $168$ . I'm pretty sure the next step is just that simple group, and the sequence is stable from there.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

227 posts

totem

#6 h Jul 18, 2015, 9:20 PM • 2 Y

Y by Adventure10, Mango247

Thanks jmerry. But precisely, how do we find $\operatorname{aut}^2 \mathbb{Z}_{15}^*$ ? Never did this before. Here for example are the automorphisms for $\mathbb{Z}_{15}^*$ . First column is auto name, second is the automorphism, third is the order of the automorphism (by composition). Since the size is 8, then we need a generator pairing of the form $\big(a,b\big)\to \big(c,d\big)$ with the order of a and c equal to 2 and order of b and d equal to 4. Take for example the generator mapping:
$(a_2,a_6)\to (a_3,a_8)$

Does this mapping lead to an element in $\operatorname{aut}^2 \mathbb{Z}_{15}^*$ . It's a bit messy; we're no longer just multiplying numbers but rather forming compositions of functions now to compute the automorphism.

Is this the way I'll have to find them? I did some work showing how to determine when generator mappings lead to automorphisms and I suspect I can extend that work to $\operatorname{aut}^2$ .

So the question now is, "find $\operatorname{aut}^2 \mathbb{Z}_{15}^*$ " explicitly. I think I can figure it out.

$\left( \begin{array}{ccc} a_1 & \{1,2,4,7,8,11,13,14\} & 1 \\ a_2 & \{1,7,4,2,13,11,8,14\} & 2 \\ a_3 & \{1,8,4,13,2,11,7,14\} & 2 \\ a_4 & \{1,13,4,8,7,11,2,14\} & 2 \\ a_5 & \{1,2,4,13,8,14,7,11\} & 2 \\ a_6 & \{1,7,4,8,13,14,2,11\} & 4 \\ a_7 & \{1,8,4,7,2,14,13,11\} & 2 \\ a_8 & \{1,13,4,2,7,14,8,11\} & 4 \\ \end{array} \right)$

Edit: It wasn't easy. I believe $\big|\operatorname{aut}^2 \mathbb{Z}_{15}^*\big|=8$ and:
$\{a_1,a_4,a_3,a_2,a_5,a_8,a_7,a_6\}\in \operatorname{aut}^2 \mathbb{Z}_{15}$
that is $a_1\to a_1$ , $a_2\to a_4$ , $a_3\to a_3$ and so on. Not 100% sure though at this point.

This post has been edited 3 times. Last edited by totem, Jul 19, 2015, 4:49 PM
Reason: add edit

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1704 posts

#7 h Jul 18, 2015, 11:34 PM • 2 Y

Y by Adventure10, Mango247

For the original problem (and more) see http://mathoverflow.net/a/94994/32216 ( $G$ doesn't even need to be finite).

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

12096 posts

jmerry

#8 h Jul 19, 2015, 5:10 AM • 2 Y

Y by Adventure10, Mango247

Of course, that means the original problem needs a correction - it applies only to non-abelian simple groups.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1704 posts

#9 h Jul 19, 2015, 8:35 PM • 2 Y

Y by Adventure10, Mango247

Yes $G$ must be non-abelian, I should have mentioned that. Of course every abelian simple group $G$ is finite cyclic of prime order $p$ , then $\operatorname{Aut}(G)\cong(\mathbb{Z}/p)^\times$ is cyclic of order $p-1$ and $\operatorname{Inn}(\operatorname{Aut}(G))$ is trivial. But $\operatorname{Aut}(\operatorname{Aut}(G))\cong\big(\mathbb{Z}/(p-1)\big)^\times$ is trivial only for $p\le3$ .

This post has been edited 2 times. Last edited by GreenKeeper, Aug 18, 2015, 3:14 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

80 posts

#10 h Jul 25, 2015, 8:40 PM • 2 Y

Y by Adventure10, Mango247

there is a condition missed in the problem: $G$ must be non-abelian.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

80 posts

#11 h Aug 2, 2015, 11:03 PM • 1 Y

Y by Adventure10

Here is my solution:
define $A=AutG,I=InnG$ , $C_{AutA}(I)=\{\alpha \in AutA: \forall \beta \in I, {\beta}^{\alpha}=\beta \}$ .
$G$ is simple non-abelian group, so $Z(G)=1$ and $G \cong G/Z(G) \cong I$ , which means
$C_{A}(I)=\{\alpha \in A: \forall \varphi_{a} \in I, {\varphi_{a}}^{\alpha}=\varphi_{a^{\alpha}}=\varphi_{a} (\Leftrightarrow a^{\alpha}=a,\forall a \in G )\}=1$ .
$I$ is characteristic subgroup of $A$ ( $\forall \alpha \in AutG, I^{\alpha}=I$ ) because:
$I$ is simple group, $\alpha \in AutG$ , $I^{\alpha}\cap I=1 \Rightarrow [I^{\alpha},I]=1 \Rightarrow I^{\alpha}\leq C_{A}(I)=1$ , which contradicts the fact $G \neq 1$ .

This post has been edited 1 time. Last edited by chenyuandong, Aug 3, 2015, 8:05 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

80 posts

#12 h Aug 3, 2015, 12:39 AM • 2 Y

Y by Adventure10, Mango247

So, $C_{AutA}(I)$ as subgroup of $AutA$ can conjugate act on these two sets $A - I$ and $A/I$ and we get two equations:
$|A|-|I|=|C_{A-I}(C_{AutA}(I))|+\sum_{\alpha \in A-I}{|C_{AutA}(I): C_{AutA}(I)\cap C_{AutA}(\alpha)|}$ ...(1)

$|A|/|I|=|C_{A/I}(C_{AutA}(I))|+\sum_{I\alpha \in A/I}{|C_{AutA}(I): C_{AutA}(I)\cap N_{AutA}(I\alpha)|}$ ...(2)

Apparently, $C_{AutA}(I)\cap C_{AutA}(\alpha)=C_{AutA}(I)\cap C_{AutA}(I\alpha)$ in the first equation.
And even further, we assert that $C_{AutA}(I)\cap N_{AutA}(I\alpha)=C_{AutA}(I)\cap C_{AutA}(I\alpha)$ :

$\chi \in C_{AutA}(I)\cap N_{AutA}(I\alpha),\alpha ^{\chi}=i_{\alpha}\alpha$
$(\alpha i)^{\chi}=\alpha ^{\chi}i=i_{\alpha} \alpha i=i_{\alpha} i \alpha ^{i}=i_{\alpha} i^{\alpha^{-1}} \alpha$
$(\alpha i)^{\chi}=(i \alpha ^{i})^{\chi}=(i^{\alpha^{-1}} \alpha)^{\chi}=i^{\alpha^{-1}}i_{\alpha} \alpha$
so, $i_{\alpha} i^{\alpha^{-1}}=i^{\alpha^{-1}}i_{\alpha},i_{\alpha} \in Z(I) \cong Z(G)=1,\alpha ^{\chi}=\alpha$
,which indicates $\chi \in C_{AutA}(I)\cap C_{AutA}(I\alpha)$ .

$\sum_{\alpha \in A-I}{|C_{AutA}(I): C_{AutA}(I)\cap C_{AutA}(\alpha)|}=\sum_{I\alpha \in A/I}{|C_{AutA}(I): C_{AutA}(I)\cap N_{AutA}(I\alpha)|}$
.............(3)

By definition, we find the fourth equation:
$|C_{A-I}(C_{AutA}(I))|+|I|=|C_{A/I}(C_{AutA}(I))||I|$ ...(4)
From these four equations, we figure out $|C_{A-I}(C_{AutA}(I))|=|A|-|I|$ which is equivalent to $C_{AutA}(I)=1$

This post has been edited 7 times. Last edited by chenyuandong, Aug 3, 2015, 1:01 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

80 posts

#13 h Aug 3, 2015, 1:01 AM • 2 Y

Y by Adventure10, Mango247

And $N_{A}(I)/C_{A}(I) \cong$ a subgroup of $Aut I$ and $(AutA) \arrowvert _{I} \leq Aut I$
$\Rightarrow (AutA) \arrowvert _{I}= (Inn A) \arrowvert _{I}=Aut I$
$\Rightarrow N_{AutA}(I)/C_{AutA}(I) \cong AutI \Rightarrow N_{AutA}(I) \cong AutI$ because $C_{AutA}(I)=1$
Finally, $I$ is characteristic subgroup of $A$ , so $AutA=N_{AutA}(I)\cong AutI \cong (Inn A) \arrowvert _{I}$
,which means $|AutA| \leq |InnA|$ , $Aut A= InnA$ , done.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a