Difference between group algebra over a field and algebra over the same field?

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?

• What do you mean by "an algebra over the same field, that is generated by the group"? – Eric Wofsey Oct 31 '18 at 1:48
• 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? – compl11112222 Oct 31 '18 at 1:51
• Again, what do you mean by "the F-algebra generated by G"? – Eric Wofsey Oct 31 '18 at 1:52
• 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"). – compl11112222 Oct 31 '18 at 2:00
• 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. – GNUSupporter 8964民主女神 地下教會 Dec 9 '18 at 22:40

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 $$G$$.