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When is there a difference between the group algebra over a field and an algebra over the same field, that is generated by a multiplicative subgroup that is isomorphic to the group? I think that for finite groups, the two notions should be the same, and that the difference occurs for some infinite groups?

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  • $\begingroup$ What do you mean by "an algebra over the same field, that is generated by the group"? $\endgroup$ – Eric Wofsey Oct 31 '18 at 1:48
  • $\begingroup$ Lets have a group G and a field F. I am wondering when is F[G] (the group algebra) different to the F-algebra generated by G? $\endgroup$ – compl11112222 Oct 31 '18 at 1:51
  • $\begingroup$ Again, what do you mean by "the F-algebra generated by G"? $\endgroup$ – Eric Wofsey Oct 31 '18 at 1:52
  • $\begingroup$ I mean subalgebra of an F-algebra B generated by elements of G, where G is a subgroup of the B* ("linear span of G"). $\endgroup$ – compl11112222 Oct 31 '18 at 2:00
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    $\begingroup$ Please don't self-delete your post. That's unfair to the answerer who has spent time and effort in answering your question. We can (and will) undelete it. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 9 '18 at 22:40
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An $F$-algebra $A$ generated by a multiplicative subgroup $G$ does not have to be isomorphic to the group algebra $F[G]$, regardless of any finiteness conditions. For a very simple example, taking $F=\mathbb{Q}$, then $A=\mathbb{Q}$ is generated by the group $G=\{1,-1\}$ as a $\mathbb{Q}$-algebra but is not isomorphic to $\mathbb{Q}[G]$.

In order to conclude that $A$ is isomorphic to $F[G]$, you need to additionally know that $G$ is $F$-linearly independent as a subset of $A$. That means exactly that the canonical homomorphism $F[G]\to A$ which is the identity on $G$ is an isomorphism, since $F[G]$ consists of formal linear combinations of elements of $G$.

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