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Let $G$ be a group and let $\mathscr{S}(G)$ denote the group of Inner-Automorphisms of $G$.

The only isomorphism theorem I know, that connects a group to its inner-automorphism is: $$G/Z(G) \cong \mathscr{S}(G)$$ where $Z(G)$ is the center of the group. Now, if $Z(G) =\{e\}$ then one can see that $G \cong \mathscr{S}(G)$. What about the converse?

That is if $G \cong \mathscr{S}(G)$ does it imply that $Z(G)=\{e\}$? In other word's I need to know whether there are groups with non-trivial center which are isomorphic to their group of Inner-Automorphisms. Are there any type of classification for such type of groups?

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Just a small comment: If G satisfies this, then the upper central series for G cannot terminate, so G cannot be nilpotent. – Tobias Kildetoft May 12 '11 at 8:52
Not really a mathematical comment, but I guess a "fun" way to state it would be : for such groups, $G/Z(G)$ is NOT centerless. – Joel Cohen May 12 '11 at 13:18
I dont know the reason for the downvote – user9413 Jun 11 '11 at 17:26
I did not downvote. However, my first thought when I saw this question was that it is trivial; since I am very used to thinking about finite groups, I assumed that $G$ is finite in this question. Of course, the point of the question is defeated if one assumes this. My guess is that someone saw this question and downvoted thinking that it was one about finite groups; perhaps it would be better to explicitly state in the beginning that it is trivial for finite groups. However, this is only my guess at why your question is downvoted. – Amitesh Datta Jun 12 '11 at 12:05
@Amitesh: Dear Amitesh, I am not blaming you. I am asking the reason to whosoever downvoted. – user9413 Jun 12 '11 at 15:37

Interesting question! It seems that there is such a group. I have no idea about a classification of these groups, myself.

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This immediately suggests a more refined question: if the natural map $G \to \mathscr{S}(G)$ is an isomorphism, does $Z(G) = 1$? – Hurkyl Aug 6 '12 at 10:41

I can help a little, maybe not if you already thought of it : there is no point in trying to find counter examples with finite groups ; for if $G \cong \mathscr S(G)$, since $\mathscr S(G) \cong G/Z(G)$, we have $G \cong G/Z(G)$, which means those two are in bijection. Since this means $|G| = |G|/|Z(G)|$, we have $|Z(G)| = 1$ and $Z(G) = \{ e \}$.

Now while looking at infinite groups it may be interesting that you are basically trying to either find a group $G$ with a non-trivial center such that $G \cong G/Z(G)$, or trying to prove that every group with a non-trivial center is such that $G \ncong G/Z(G)$. My first approach would be to try some nasty groups as counter-examples rather than trying proofs of the second choice...

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@mac - yet with this example you can create an infinite family of groups G satisfying $G \cong G/Z(G)$ by taking the direct product with any group having trivial center. Since you cannot possibly classify all groups having a trivial center, it is hopeless to expect a classification of the class of groups proposed by Chandru.

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