Let $H$ be a torsion-free abelian group and let $\mathbb{K}$ be a field with two elements. I would like to ask the following question:
Is the group of units of the group algebra $\mathbb{K}[H]$ isomorphic to $H$ ?
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$\begingroup$ No, obviously not, because $\mathbb{K}^\times \leq (\mathbb{K}[H])^\times$. $\endgroup$– Johannes HahnCommented Feb 2, 2016 at 19:37
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$\begingroup$ oops thanks, I will change my question. $\endgroup$– OfraCommented Feb 2, 2016 at 19:42
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$\begingroup$ Are we assuming $H$ to be finitely generated or not necessarily? In the finitely generated case the answer is yes. $\endgroup$– Yonatan HarpazCommented Feb 2, 2016 at 20:31
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2$\begingroup$ I just realized it doesn't matter. If $x \in \mathbb{K}[H]$ is invertible then there exists a finitely generated subgroup $H' \subseteq H$ such that $\mathbb{K}[H'] \subseteq \mathbb{K}[H]$ contains both $x$ and $x^{-1}$, and so we may as well assume that $H$ is finitely generated. The answer hence is yes. $\endgroup$– Yonatan HarpazCommented Feb 2, 2016 at 20:44
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2$\begingroup$ It's indeed a classical conjecture of Kaplansky (en.wikipedia.org/wiki/Kaplansky%27s_conjecture) that if $K$ is a field , the units of $KG$ are the multiples of elements in the standard basis, and is well-known and easy when $G$ is abelian. In particular if $K$ has 2 elements, the obvious homomorphism $G\to KG^\times$ is an isomorphism (which is the right statement, just saying that they are isomorphic is weaker and much less interesting). $\endgroup$– YCorCommented Feb 2, 2016 at 21:16
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