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Claim: Let $G$ be a group. Prove that $G \cong\mathrm{Inn}(G)$ if and only if $Z(G)$ is trivial.

Could anyone offer a hint on proving this claim just using simple properties of group isomorphisms? (i.e., not using the fact that the quotient group $G / Z(G)$ is isomorphic to the group of inner automorphisms of $G$.)

EDIT:

It turns out that this Claim is false as stated. There are several counterexamples, several of which are provided in answers below, for the case of G being infinite. However, the claim holds for finite groups.

luke
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    But why not use the isomorphism you mentioned? Is it because of the arrangements of the book you are reading? Or is it because you have a feeling that this could work? Thanks for any explanation. – awllower Mar 08 '13 at 05:28
  • The text I am reading presents this problem before its discussion of quotient/factor groups, so I am trying to determine this more basic approach to solving it. – luke Mar 08 '13 at 05:33
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    This is probably false. The correct statement is not about $G$ and $\text{Inn}(G)$ being isomorphic but about a specific map between them (namely the map $g \mapsto (x \mapsto gxg^{-1})$) being an isomorphism. You don't need to know anything about quotient groups, as such, to solve this version of the problem: you just need to determine when this map is injective. – Qiaochu Yuan Mar 08 '13 at 05:34
  • @zach, I think there's a very slim chance this can be properly proved without at least knowing quotient groups...which, btw, are usually studied way before center subgroup, automorphism group and etc. – DonAntonio Mar 08 '13 at 05:37
  • @QiaochuYuan, I think the wording of the problem is correct. – DonAntonio Mar 08 '13 at 05:39
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    @Don: what is your argument for the implication that if $G$ is isomorphic to $\text{Inn}(G)$ then $Z(G)$ is trivial? – Qiaochu Yuan Mar 08 '13 at 05:40
  • Well, as stated below: I missed the iff thingy...One direction is true, though. – DonAntonio Mar 08 '13 at 05:43
  • @zach With respect to your edit, it holds for all Hopfian groups. Finite groups are just special examples of Hopfian groups. – user1729 Mar 17 '13 at 19:35

3 Answers3

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The claim is false as stated. groupprops has a counterexample.

The correct claim is specifically about the natural map $G \to \text{Inn}(G)$, and as I mentioned in my comment above you don't need to know anything about quotient groups to prove the correct claim: you just need to determine when this map is injective.

Qiaochu Yuan
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Hints: if $\,\phi_a\,$ denotes the inner isomorphism determined by $\,a\,$ , i.e. $\,\phi_a(x):=axa^{-1}\,$ , then

(1) Check that $\,F:G\to\operatorname{Aut}(G)\;,\;\;F(a):=\phi_a\,$ , is a homomorphism ;

(2) Prove $\,Z(G)=\ker(F)\,\,,\,\,\,\operatorname{Inn}(G)=\operatorname{Im}(F)$

(3) Apply the first isomorphism theorem

Added: As already noted in the comments, this only shows that $\,Z(G)=1\Longrightarrow G\cong Inn(G)\,$

DonAntonio
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    This does not prove the claim as stated. – Qiaochu Yuan Mar 08 '13 at 05:36
  • Unless I painfully misunderstood something, I think it does: it shows that $,G/Z(G)\cong Inn(G),$ and from here the claim follows at once...Of course, it uses quotient groups. – DonAntonio Mar 08 '13 at 05:41
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    No it doesn't. From here it follows that the natural map $G \to G/Z(G)$ is an isomorphism iff $Z(G)$ is trivial. This argument doesn't say anything about the possibility that $G$ and $G/Z(G)$ might not be isomorphic by this map but might be isomorphic by some other map. – Qiaochu Yuan Mar 08 '13 at 05:42
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    The other direction follows if, for example, $,G,$ is hopfian, but I severely doubt this is what was intended... – DonAntonio Mar 08 '13 at 05:45
  • @DonAntonio ...and only if? (I wonder.) – Alexander Gruber Mar 08 '13 at 06:08
  • I already answered in the above comments (below the question): I missed the iff part. The other direction is true with further conditions, one of which can be hopfian and, apparently, someone already posted an answer with a link that shows that the iff claim is false. – DonAntonio Mar 08 '13 at 08:16
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A somewhat simpler and better known (if maybe not for this property) counterexample against the claim than the one cited by Qiaochu (but it is very similar). Take $G=O(2,\mathbf R)$, the orthogonal group of the plane. Choose a line $L$ though the origin, then there is a surjective morphism of groups $G\to G$ that sends every rotation $\rho$ to $\rho^2$, and fixes the reflection $\sigma_L$ with respect to $L$. Every other reflection is of the form $\sigma'=\rho\sigma_L$ maps to $\rho^2\sigma_L$, which is the reflection wit respect to the line $\sigma'(L)$ (which shows that all reflections are in the image); it is easily checked that this is a morphism of groups. The kernel of this morphism has two elements, the identity and the half-turn rotation, which also form the center of $G$. Therefore by the first isomorphism theorem $G\cong G/Z(G)\cong\operatorname{Inn}(G)$ even though $Z(G)$ is not trivial.

Marc van Leeuwen
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