I am studying Linear Algebraic Groups on the book of T.A.Springer. I have some questions:
- Why a finite group can become an algebraic group?
- Could you give an example of a group that can not become an algebraic group?
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1Every finite subset of say the affine line, is a closed affine set. – Angina Seng Dec 21 '17 at 15:34
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- Try this first for the finite permutation group. 2. Take for instance the Higman group.
– Moishe Kohan Dec 21 '17 at 16:20
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I will assume that we have as a given that the group $\operatorname{GL}_n$ is an algebraic group for every $n\in\Bbb N$.
Then, note that every subgroup of $\operatorname{GL}_n$ which is a Zariski-closed subset (sometimes also referred to as a closed subgroup) is again an algebraic group: It is an algebraic variety because it is a closed subset of the variety $\operatorname{GL}_n$ and since the group operations are inherited, they are also morphisms of varieties.
The symmetric group $\mathfrak S_n$ is a subgroup of $\operatorname{GL}_n$ in the form of permutation matrices and every finite group is a subgroup of $\mathfrak S_n$ for some $n\in\Bbb N$, hence every finite group is a subgroup of $\operatorname{GL}_n$ for some $n\in\Bbb N$.
Jesko Hüttenhain
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Is it possible to see what $I$ and $ k[x_1,\ldots,x_n]/I$ will be, with say $S_3 \subset GL_6$ ? – reuns Dec 22 '17 at 10:50
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Concerning $I$, that might be tricky and I do not know off hand, but the coordinate ring of a finite set of points is always just a product of as many copies of $k$ as there are points. – Jesko Hüttenhain Dec 23 '17 at 09:59
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The rational map representing the multiplication by $g$ will be just a permutation of the $x_m$ ? – reuns Dec 23 '17 at 18:04