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I've read not all finite groups are automorphisms of a finite group, but I was reasoning (probably wrongly) like this:

Let's take $\Bbb Z_4$ and try to figure out some automorphisms out if it.

We can define $\Phi(\rho) = \rho^2$, and by this we'll get $\Bbb Z_2$ as automorphism group.

Since the modularity of $\Bbb Z_4$ any power of $g$ which will be $2(1+2n)$ will give the same $\Bbb Z_2$ as result.

Now, if take $\Phi(\rho)=\rho$ i.e. the $\Phi$ as the identity function, we'll have the other automorphism, i.e. $\Bbb Z_4$ itself. This is true for any $1+2n$.

Notice some powers are excluded, like $\Phi(\rho)=\rho^4$, since would send all elements to to the identity, violating the inejctivity of the homomorphism.

So, basically here I see two groups, i.e. $\Bbb Z_2$ and $\Bbb Z_4$.

So, I was guessing: if we take the identity function we will always have the same group as itself automorphism.

So, if this holds, every finite group is the automorphism of a at least one finite group, i.e. itself.

Where am I wrong in my reasoning, please?

Incidentally: I came across this since the book I'm reading defined $\Phi(\rho)=\rho^3$ as the function to be used to compute the $\Bbb Z_4$ automorphism, but using it I stumbled quickly in the whole group, so I realized by myself the right solution is using $2$ as power.

Tx in advance.

Shaun
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riccardoventrella
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    $\rho\mapsto\rho^2$ is not an automorphism of the cyclic group with four elements: it's not injective. Moreover, you're also misunderstanding the term "automorphism group" - the (there's only one) automorphism group of a group $G$ is the group $H$ of all automorphisms of $G$, under composition. – Noah Schweber Jul 16 '22 at 16:17
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    For example, the automorphism group of $\mathbb{Z}/4\mathbb{Z}$ (your "$Z_4$") is just $\mathbb{Z}/2\mathbb{Z}$: there are two automorphisms, the trivial one $x\mapsto x$ and the map $x\mapsto -x$. (Strictly speaking that only tells us that $\mathsf{Aut}(\mathbb{Z}/4\mathbb{Z})$ has two elements, but there is only one group with two elements so that's enough.) – Noah Schweber Jul 16 '22 at 16:19
  • So, using the "power notation" used in the book, which $n$ will be the power realizing $Z_2$? – riccardoventrella Jul 16 '22 at 16:27
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    I don't understand what you're asking - there is no "realizing" in this context. $\mathbb{Z}/4\mathbb{Z}$ has two automorphisms, namely $$\alpha:x\mapsto x\quad\mbox{and}\quad \beta:x\mapsto x^3.$$ The set ${\alpha,\beta}$ forms a group under composition; its identity element is $\alpha$, and it is isomorphic to $\mathbb{Z}/2\mathbb{Z}$. That's all there is to say. (Note that it's potentially misleading to use exponential notation here; "$x\mapsto 3x$" is much clearer in my opinion.) – Noah Schweber Jul 16 '22 at 16:39
  • The following should answer your question: https://math.stackexchange.com/q/253936/104041 – Shaun Jul 16 '22 at 16:42
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    I agree with @NoahSchweber as the group is abelian so additive notation is more natural and common. This is motivated by the fundamental theorem of finite abelian groups. You may also be interested in inner automorphism and outer automorphisms to better your understanding of automorphism groups. – CyclotomicField Jul 16 '22 at 17:04
  • @NoahSchweber, I start realizing now the big misunderstandingI had on automorphisms, tx. However I'm still struggling to understand how $\alpha$ can act as indentity. I mean, it's clear it's an identity from the G group of view, since sending every element to itself, but I cannot see how it's treated as an identity in the ${\alpha, \beta}$ group. May you promote your comment as an answer, enriching it by some example of those 2 elements multiplication? Tx in advance. – riccardoventrella Jul 16 '22 at 17:22
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    @riccardoventrella Think about how composition of functions works: $(\alpha\circ\beta)(x)=\alpha(\beta(x))=\beta(x)$, so $\alpha\circ\beta=\beta$. Similarly, $\beta\circ\alpha=\beta$. – Noah Schweber Jul 16 '22 at 17:29
  • @CyclotomicField, so far I dealt only in building automorphisms from conjugation, i.e. forming inner automorphisms. But here we have an abelian group, which has a trivial inner automorphism group. I was interested in building manually the group here, like for other groups like $S_3$ or $K_4$ – riccardoventrella Jul 16 '22 at 17:41
  • So for example $x\mapsto5x$ Is a good automorphism, but cannot be part of any group, right? I mean, among the infinite good automorphisms, only those 2 are forming a group, correct? Or it''s because produces the same result as the identity? – riccardoventrella Jul 16 '22 at 18:19

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No cyclic group, $C_n, n\gt1$, is its own automorphism group, because generators go to generators, and the automorphism group thus has order $\varphi (n)$. Since $\varphi (n)\not=n$, the groups are different.

Neither is $\Bbb Z$, which only has $2$ automorphisms.

In fact, this does happen though. If the center and outer automorphism group are trivial, this occurs, and the group is called complete.

For instance, $S_n, n\not=2,6$.

Just use the isomorphism between $G/Z(G)$ and $\rm{Inn}(G)$.

$S_6$ is not complete, because it has an outer automorphism.

One more result: the automorphism group of a non-abelian simple group is complete.

So, for instance, $A_n,n\ge5, n\ne6$. In these cases, $\rm{Aut}(A_n)\cong S_n$.

$\rm{Aut}(A_6)\cong P\Gamma L(2,9)$. It's the "projective semi-linear group of degree $2$ over the field of $9$ elements". And it's complete.

There are groups that are not complete, but still equal to their automorphism group. For instance $D_8$. It isn't complete because it's center is $\Bbb Z_2$. (It also has outer automorphisms. All the dihedral groups do except $D_6$.)

calc ll
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