Automorphisms Puns

I was listening to Morph too much. And googling just to kill the time too much too.

Found N. Bourbaki mentioned in proofs for automorphisms – for instance:

https://preview.redd.it/rochylcc7sa61.jpg?width=800&format=pjpg&auto=webp&s=7abd93acc22580115a3a0596d3df10b7b7557acd

https://mathoverflow.net/questions/90964/lie-algebra-admitting-some-hyperbolic-automorphism-is-nilpotent

of which I understand nothing – I’m not good in Algebra (nor in English – quick apologies for people enjoying correct grammar – English is my second language, so please skip as there’s nothing of importance here :) it’s just this unresolved duality of dualism keeping me up

And the ones and zeros were bothering me. I tried looking for less advanced Algebra and found identity matrix and identity morphisms.

where I is the identity matrix (which is trivially involutory); R is an identity matrix with a pair of interchanged rows; [from Wikipedia of course]

Automorphism is a symmetry of the object, and a way of mapping the object to itself while preserving all of its structure.

The identity morphism (identity mapping) is called the trivial automorphism.

An identity relation or identity map or identity transformation, is a function that always returns the same value that was used as its argument.

the identity matrix is invertible—with its inverse being precisely itself [end of Wikipedia quotes]

Not sure if it makes any sense, but since many are now making wild guesses about the sixth album – my guess would be it might touch the subject of identity. I know it is more a wishful thinking from what I am personally working through just recently, but I just wanted to put it out there.

Also people in dema are described as devoid of any trace of an identity in Clancy’s letter. Maybe it could be the next step – not only fight and not only escape but work on what we allow to define ourself and why. I have no idea how to put in an understandable way – sorry.

I’m not sure if it was already discussed? If anyone can point me where I would appreciate. I know this is a long shot and based off practically nothing, but in case someone would like to have look into that - let me know what you think?

I have a question about the following exercise:

Let C be a full subcategory of topological spaces containing the singleton space {.} and the Sierpinski space {0,1} where {1} is open and {0} is closed. First, show FS is isomorphic to S. Then, show there is a unique natural isomorphism \alpha: U -> UF, where U:C -> Set is the forgetful functor. Finally, if C additionally contains a set X in which not all unions of closed sets are closed, then \alpha_S : US -> UFS is continuous, and deduce that F is naturally isomorphic to the identity functor.

My question is: In part III, is the condition that C contains such a set X necessary for the conclusion. I came up with an argument that doesn't require it, and I'm wondering if I've gone wrong.

as a postscript I'll include my argument as well for those interested enough to read it. I first showed in part II that both U and UF are naturally isomorphic to the functor Hom( . , _) in Set, and conjoining the canonical isomorphisms gives an explicit description of (\alpha_s)^-1 as x |--> ( f: . ->F^{-1} Id_FS (x) ) |-->Uf( . ) =F^{-1} Id_FS (x). But then we may note that since F is an automorphism and Id_s is the unique invertible endomorphism of S then F^{-1} Id_FS = Id_s hence (\alpha_s)^-1 is the identity, and is continuous.

Edit: I have actually discovered it must be a necessary condition, but still am unsure what is wrong in my proof. If C is just {.} and S, then there is the automorphism F which swaps the maps ( . -> 1) and (. -> 0), but still takes the identity of S to itself. Then there cannot be an natural isomorphism to the identity functor.

Hello! I have been stuck on the following problem for a while. Could you please give me a hint? Ideally, I would like to solve the problem using the hint. Thank you so much!

Find all automorphisms of order $3$ of $Z_{91}$ (Hint: How can $Z_3$ act non-trivially on $Z_{91}$?)

Are there any automorphisms of order $5$? 

\vspace{1cm}

The only stuff I was able to figure out is this:

Let $G = Z_{91}$. I know that $\text{Aut}(G) \cong (\mathbb{Z}/91\mathbb{Z})^{\times}$ which has $\varphi(91)=72$ elements, so I can already tell that $Z_{91}$ has no automorphism of order $5$. 

Since $\text{Aut}(G)$ is abelian, the set of all elements of $\text{Aut}(G)$ which satisfy $x^3=1$ form a subgroup.

Here's what's in my textbook:

"Determine all automorphisms of the field of rational numbers.

Solution: Let f : Q -> Q be an automorphism of fields. It means that f is an automorphism of the group (Q, +), therefore, f(x) = x * f(1). Because f(1)=1 it means that f(x)=x for any rational x, the only automorphism of the field Q."

Hold up. I didn't understand this part: therefore, f(x) = x * f(1). What does this have to do with the fact that it's an automorphism on the group (Q, +). From that it should result that f(x+y) = f(x) + f(y). Instead, they used multiplication, which should result from the fact that it's an automorphism on the group (Q, ), not (Q, +). And even then, it would result that f(xy) = f(x) * f(y), why is f(x) = x the only function doing that? I don't understand this proof/solution at all.

Hi all,

I've worked on what looks like a rather interesting problem in Galois Theory for hours now, and haven't come up with a solution.

I'll define some notation before I state my problem. For a field L, let Aut(L) denote the group of automorphisms on that field. For a subgroup G of Aut(L), denote by L^G the fixed field of L under automorphisms from G (i.e. the elements of L that are mapped to themselves by all automorphisms in G).

The actual problem: I'm having trouble finding a field L with a subgroup G of Aut(L) such that the field extension L/L^G is *not* galois.

So far, I've figured out that the subgroup G cannot be finite and the extension L/L^G cannot be algebraic, otherwise a standard proof for separability and normality can be applied (happy to elaborate if needed). Other than that, I've not really come up with any ideas. Any suggestions are welcome.

I understand that a group automorphism is an isomorphism from a group to itself. However, there are a few aspects of group automorphisms that make the topic a bit hazy to me:

-What are the motivations of automorphisms and automorphism groups?

-How does one construct automorphisms and automorphism groups?

-What is an example of an outer automorphism?

Also, on a somewhat side note, one of the main resources I have been using to aid my study of group theory, the YouTube channel Ben1994, often suggests that viewing groups as the set of permutations on a set can be helpful and insightful, but I don't understand why groups in general can be viewed as compositions of set permutations and vice versa.

Hi can someone please help me with the following question. I am not to sure how I would do it.

Verify that f1=0 1 2 3 4 0 2 4 1 3

                 f2= 0 1 2 3 4
                         0 3 1 4 2 
  
                 f3= 0 1 2 3 4 
                         0 4 3 2 1 

Are all Automorphisms of Z5 (integers)

[; \alpha :\mathbb{Z}_n\to\mathbb{Z}_n;] by [; s\to sr;] where [; r\in U(n);] is an automorphism.

Having some issue showing this is a homomorphism:

Since [; gcd(n,r)=1;] then there exist integers x,y such that [; xr+yn=1;]

[; \alpha(st)=str=str(xr+yn)=str^2x+stryn=sr(trx+tyn)=sr(trx)=\alpha(s)\alpha(t)x;]

But I want to show that [; \alpha (st)=\alpha(s) \alpha(t);]

Suppose ϕ is an automorphism (other than the identity function) of a group G, and let ϕⁿ be ϕ composed with itself n times. Is it possible for ϕ to have the property that, for any a, b, c ∈ G:

there exists an integer n such that

ϕⁿ(a) = b

implies that there exists an integer m such that

ϕᵐ(ac) = bc

What kind of automorphisms have this property? Does anyone have an example (if it's possible at all)?

edited for clarity

Q_8 is the quaternion group and I'm just looking for an example of one that is not an outer automorphism, in general I struggle with finding examples because I have no clue where to begin. Any help appreciated!

Not sure what standard notation is (if any), but by U(16) I mean the group of natural numbers less than 16 and relatively prime to 16, with multiplication mod(16).

My answer so far:
f(x) clearly maps elements of U(16) back to U(16) due to closure under multiplication.

It's not hard to show that f(x) is operation preserving (because U(16) is Abelian).

But I need to show that f(x) 1-to-1 (or onto- either will do since U(16) is finite).
I can take all the elements of U(16) and cube them mod16 to show that there are no duplicates. That doesn't feel very slick though.
Is there a better way? This has had me stumped for days. Thanks!

suppose i was given a question "How many automorphisms of D4 are there?" would this be different for different groups (i.e. Zn, U(n), Dn, etc.) and is there a general way of finding the answer for each of these groups. from what i understand, and automorphism is the mapping of a generator in G to the generator and also read that if G has order m then

|Aut(G)|=ϕ(m) where ϕ(m) is Euler's function.

although i dont think this is true for D4 since the number of automorphisms in D4 is 8. (is this right?)

How do you find the automorphism group of a direct product?

Like suppose I have Z/mZ x Z/nZ What is Aut(Z/mZ x Z/nZ)?

Some interesting questions that arose when I was talking with a friend. The last one we have yet to solve.

1. Can you find a finite group G, with subgroup H, so that Aut(G) is much smaller than Aut(H)?

2. Consider G=S_n +S_n +..+S_n (direct sum k times, or direct product since they aren't abelian is maybe better). (think of n as fixed, and k growing to infinity). Prove that you need at least k generators to generate G.

3. Is there a finite group G so that lim n-> infty log( |Aut(G^n )| ) /(nlogn) <infty (limsup if you prefer is also okay) (where G^n is direct product of n copies of G)

For such a general concept like automorphisms you would think they would use it for more than that

For every group, an automorphism is an isomorphism from the group to itself. For example, G -> G: g -> g^(-1) is an automorphism if G is abelian, since it preserves the group structure, is bijective, and is from G to G.

It can easily be shown that the set of automorphisms of a group G (hereafter denoted as Aut(G)) is itself a group under composition. Furthermore, for some groups (e.g. S3), Aut(G) is isomorphic to G, and for other groups (e.g. Z), Aut(G) is not ismorphic to G.

Does there exist some group G such that Aut(G) is not isomorphic to G, but Aut(Aut(Aut(...(Aut(G))...))) is isomorphic to G?

Since I don't know the difficulty of this problem, since it's not yet solved, I've marked it as "Hard". However, I wish we had an "Unsolved" flair for questions like these. Perhaps on a grey background, like the old "Impossible" flair.

Hi,

Suppose I have an automorphism f:G--->G where G is finite. Suppose f has order t. Suppose that for some g in G, k is the smallest positive natural number that satisfies f^k (g)=g. Does k divide t? And if it does, could someone explain why?

Thank you

I'm watching a youtube series on undergraduate abstract algebra and the speaker (Benedict Gross) at one point makes a comment that without any calculation at all, we know Aut(Klein-4) is isomorphic to S3 because Aut(Klein-4) permutes the 3 natural objects in the klein 4 group (everything besides the identity).

In a later video he states something similar about 2x2 matrices over the field Z/2Z - that we know at first glance GLn(Z/2Z) is isomorphic to S3 because those matrices permute the 3 non-identity vectors in (Z/2Z)x(Z/2Z), which are (1,0), (0,1), and (1,1).

I see the point that the automorphisms do indeed permute the objects in question. What boggles me is, how do we know that, with these two groups, every such bijection is an automorphism?? In other words, I see that the set of bijections on the Klein-4 group sending the identity to the identity is isomorphic to S3. I see that the set of bijections on (Z/2Z)x(Z/2Z) sending (0,0) to (0,0) is isomorphic to S3. But how do we know that every such bijection is an automorphism, short of manually checking each one?

For instance, take the klein-4 group as {e,a,b,c}. Sure, there is a bijection for every possible permutation of a, b, and c. Take the bijection e->e, a->b, b->c, c->a.

That's a fine bijection, but how do we know that every such bijection that we can think of is automatically also a homomorphism? It seems logically possible that some such bijection could fail to preserve group multiplication.

It occurred to me that just like we can decompose finite permutations into disjoint cycles, we can decompose permutations on infinite sets in the a similar way. Except we must allow:

1. Infinite cycles of the form: [;( \hdots \ a_{-2} \ \ a_{-1} \ a_0 \ a_1 \ a_{2} \ \hdots );].

2. "Composition" of an uncountable amount of cycles.

Is this used anywhere?

PS: It reminds me of the usual proof of the Schröder–Bernstein_theorem, though it has some clear big differences. There you have 2 functions between 2 sets, and here you have 1 function on 1 set.

Suppose we have some group G_0 and we consider all the Cayley graphs that can be made by G_0's minimal generating sets. Then for each Cayley Graph produced, we can consider it's automorphism Group and repeat this process for each new group produced.

Is there anything we can say about the set of all groups that can be produced this way given some group G?

I'm trying to find a decomposition of automorphisms on a group G. For any inner automorphism 𝜙*a(x) = axa^(-1) it's pretty simple to decompose this into a left multiplication and an inverse right multiplication as follows. If 𝜏a(x) = ax and 𝜓a(x) = xa, then 𝜙a(x) = (𝜏a* ∘ 𝜓*a^(-1))(x) = (𝜓a^(-1) ∘ 𝜏a*)(x) I'm trying to find the same type of decomposition for an arbitrary 𝜙 in Aut(G)

More specifically, for an arbitrary 𝜙 in Aut(G), I'm looking for 𝜏 and 𝜓 such that neither are automorphisms on G and 𝜙(x) = (𝜏 ∘ 𝜓^(-1))(x) = (𝜓^(-1) ∘ 𝜏)(x)

Is this possible, if so are there any resources/papers on this kind of decomposition?

I recently saw this post from Ed Pegg on Math Stack Exchange about integral graphs with trivial automorphism groups. I'm interested in trying to construct smaller such graphs - at the very least, I'd like to learn about the machinery involved. What are some known results on the relationship between the automorphisms of a graph and the spectrum of the graph? I'd appreciate any explanations and / or references. Thanks!

Let a = sqrt(1+sqrt(3)) b = sqrt(1-sqrt(3)) E = Q(a,b)

I have shown that E is a galois extension and that the Galois Group G=Gal(E:Q) has order 8. I am now asked to determine the elements of G. I know that they are automorphisms completely determined by a and b. How should I proceed with this question?? i.e How do I explicitly determine what these 8 automorphisms are??
Thanks

I've got the question; Let G be a group and let H (be normal in) G. Let ro belong to Aut(G). Show that G/H is isomorphic to G/ro(H).

So far I've taken theta to be the canonical map from G to G/H, ker(theta) = H, and then by the first isomorphism theorem im(theta) is isomorphic to G/ker(theta) = G/H. Now I just need to show that G/ro(H) = im(theta) but I'm unsure how exactly to do that, and what the Automorphism does here.

Many thanks for any help!

Has anyone made the connection between andre weil + automorphism + morph + andre the bishop and the insecurities he represents? I'm assuming someone has but I can't seem to find a post/discussion about it so can someone direct me so I can read up on whoever's theory it is? I probably just didn't look hard enough so my apologies. I'm also not really good enough at math to understand automorphism and automorphic functions and representations so I'm hoping someone has come up with a theory who is lol

This week we have the usual (solutions, video links, problem set) and a link for the next video in the prerequisite playlist.

We consider further examples of automorphism groups. First we find the automorphism group of A4, and we note the Conjugation Rule for permutations. Then we have Aut(Z/n), leading to the Euler totient function and Fermat's Little Theorem. In turn, these tools are applied to find Inn, Aut, and Out for the dihedral groups (advanced material).

Class Page

Solution Set 7 - GT10/11. Non-Isomorphic Groups, etc.

Video - GT11.1. Automorphisms of A4

Video - GT12. Aut(Z/n) and Fermat's Little theorem

Video - GT12.1. Automorphisms of D_2n

Problem Set 8 - GT11/12. More Automorphisms

Regular: 1-3, 6-10; Advanced: 4, 5, 9(d), 11

For the prerequisite video, we consider binary relations on sets. Of particular interest for group theory are equivalence relations and partitions.

Prerequisite Video - BM7. Binary Relations