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Let $G$ be an infinite countable p-group. Is it true that $G\otimes G$ or $G\wedge G$ are also p-groups? (where G acts on itself by conjugation). For simplicity, you can consider that $G=[G,G]$, and then the above groups are isomorphic to the Universal Schur Cover.

I am very frustrated, because a lot of websites and papers say yes, but no one writes a clear proof (and I doubt the answer is yes). And some actually impose conditions, like G is locally finite. But I am interested in general. It might be because for some people, p-group means automatically finite? For example: http://groupprops.subwiki.org/wiki/Tensor_product_of_p-groups_is_p-group

Going to the reference, Ellis' paper: http://www.sciencedirect.com/science/article/pii/0021869387902493

we see that the theorem says yes. But he actually sends the reader to R. BROWN AND J.-L. LODAY, Van Kampen theorems for diagrams of spaces. http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.145.6183&rep=rep1&type=pdf

At page 316 they just say it is true, but I have no clue why it would be for infinite groups.

Also, the article http://ac.els-cdn.com/S0021869311006843/1-s2.0-S0021869311006843-main.pdf?_tid=541b0d28-6046-11e5-8582-00000aab0f27&acdnat=1442829268_d25fb517522e5b31af167da3ac83faf8 states at page 348, paragraph 2 that this is true, and sends the reader to some other papers. None of them look like actually do the general case: infinite countable p-groups G.

Can someone clarify this to me, with a proof or a counterexample?

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  • $\begingroup$ Brown and Loday invoke a couple of exact sequences, rather than "just say it is true". For example, $G\otimes G$ is an extension of $\nabla(G)$ by $G\wedge G$; $\nabla(G)$ is a quotient of $\Gamma(G^{\rm ab})$, the Whitehead quadratic functor; don't we know that if $G$ is a $p$-group, then so is $\Gamma(G^{\rm ab})$, hence so is $\nabla(G)$? And $G\wedge G$ is an extension of $H_2(G)$ by $[G,G]$, and both are $p$-groups when $G$ is a $p$-group... $\endgroup$
    – Arturo Magidin
    Commented Sep 22, 2015 at 19:39
  • $\begingroup$ I agree, but I got stuck at: why is $H_2(G)$ a p-group? I didn't find any references anywhere for the infinite case, and I couldn't prove this by myself. To be honest, that's what I actually wanted from the very beginning, but I decided to write the post in terms of $G\otimes G$ rather than $H_2(G)$. $\endgroup$
    – Alin Galatan
    Commented Sep 22, 2015 at 19:53

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