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I was wondering if there was a way to see any group $G$ as a subgroup of the inverse group of a ring, $G \leq A^\times$?

Thanks!

Ottavio
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Maenwe
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    Yes. Take the group ring over your favorite field. – ahulpke Dec 27 '20 at 14:52
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    For a finite group with $n$ elements it suffices to say that $G$ is a subgroup of $S_n$ which is a subgroup of $M_n(\Bbb{R})^\times$, sending the permutation $i\to \sigma(i)$ to the matrix $M_{ab} = 1$ if $\sigma(a)=b$, $M_{ab}=0$ otherwise. When the group is infinite it works the same way replacing $M_n(\Bbb{R})$ by the endomorphism ring of an infinite vector space whose basis has the same cardinality as $G$. The group ring $\Bbb{R}[G]$ is the subring generated by those matrices/endomorphisms. – reuns Dec 27 '20 at 14:59
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    The question in the title is much more interesting than the one you actually pose. – ancient mathematician Dec 27 '20 at 15:46
  • You've asked two different questions in the title and body (possibly unintentionally), and the title question is harder, has a different answer, and is also a duplicate. – Qiaochu Yuan Dec 27 '20 at 18:57

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Consider the group algebra $\mathbb{C}G$ which consist of the formal linear combinations of the form $$\sum_{g \in G}\alpha_g g$$ and where the algebra operations are the obvious ones. Note that we have an obvious inclusion $G \hookrightarrow \mathbb{C}G$. Then $G \subseteq(\mathbb{C}G)^\times$. Note however that in general the latter inclusion can be strict.

J. De Ro
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  • I think the algebra operations may not be so obvious to people who have never seen them before. To be explicit, we define $(\sum_{g\in G}\alpha_g g)\cdot (\sum_{h\in G}\beta_h h) = \sum_{g, h \in G}\alpha_g\beta_hgh$. Collecting like terms, we can put this into the form of a linear combination of elements of $G$, equal to $\sum_{g\in G}\big[(\sum_{x,y\in G: xy = g}\alpha_x\beta_y)g\big]$. – Sebastian Monnet Dec 27 '20 at 15:23
  • @Qwertiops Thanks for your comment. We get used to things when we use them a lot and we forget that some things may not be obvious :) – J. De Ro Dec 27 '20 at 15:48