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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.2011-05-12
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    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.2011-05-12
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    I dont know the reason for the downvote2011-06-11
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    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.2011-06-12
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    @Amitesh: Dear Amitesh, I am not blaming you. I am asking the reason to whosoever downvoted.2011-06-12

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