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If $G$ is a discrete torsion-free group, can its (reduced or full) group C-star algebra contain non-zero quasinilpotent elements? I've seen various examples in the group von Neumann algebra setting (usually in the context of finding quasinilpotent generators of II-1 factors) but if I remember correctly these usually don't belong to the reduced group C-star algebra.

Can we get such examples for free groups, for instance?

(My reason for assuming torsion-free is that if $G$ contains a finite non-abelian subgroup $H$, then the C-star algebra generated by H will contain a non-trivial matrix algebra and hence will contain nilpotent elements.)

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  • $\begingroup$ A question on nilpotency: Is there a $C^{*}$ algebra such that each nilpotentce element has all its roots: that is for every nilpotent $a$ and $n\in \mathbb{N}$ there is $b$ with $b^{n}=a$? $\endgroup$
    – Ali Taghavi
    Commented Nov 11, 2014 at 22:06
  • $\begingroup$ @AliTaghavi Surely this should be asked as a separate question? $\endgroup$
    – Yemon Choi
    Commented Nov 11, 2014 at 22:14

2 Answers 2

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Kaplansky showed that every non-commutative C*-algebra contains a non-zero nilpotent element. I don't have a reference I'm afraid.

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  • $\begingroup$ Ah, so I should have looked harder before asking, I guess. Thanks! $\endgroup$
    – Yemon Choi
    Commented Mar 29, 2013 at 23:18
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    $\begingroup$ For a proof for the above statement see, Page 110 (Proposition II.6.4.14) of Blackadar's book "Operator algebras". $\endgroup$
    – user23860
    Commented Mar 30, 2013 at 13:12
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Prompted by Douglas Somerset's answer to look harder, I've found a paper of Behncke that also gives what I need, according to MathSciNet, by proving a stronger result:

MR0283582 (44 #813) Behncke, H. Nilpotent elements in group algebras. (Russian summary) Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 19 (1971) 197-198.

"The aim of this note is to prove the following theorem: The group algebra $L^1(G)$ of a locally compact group $G$ has nilpotent elements if and only if $G$ is non abelian."

(I haven't yet got to the library to check the proof; in particular, I don't know if it bootstraps from Kaplansky's result as mentioned in the other answer, or mimics the proof in a different setting, or uses something more closely related to the group structure.)

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