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I am currently trying to go through Humphrey's Linear Algebraic Groups and am stuck on a problem that sounds deceivingly simple but can not seem to figure out (Problem 5 from Section 7). I have spent a fair amount of time on it and expect it to be something silly that I just am not seeing. The problem is:

A closed subset of an algebraic group which contains e and is closed under multiplication is a subgroup

  • http://math.stackexchange.com/questions/68726/a-closed-subset-of-an-algebraic-group-which-contains-e-and-is-closed-under-tak – Marci Jul 08 '13 at 17:57

1 Answers1

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Hint: Suppose $H$ is a closed subset of $G$ containing $e$ and which is closed under multiplication. Fix an element $h\in H$, and consider the sets $h^n\cdot H\subset H$.

bradhd
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