Why is $a^{\# G}=e$ where $a$ is any element of group $G$ and $\# G$ is the number of elements and $e$ is the identity element?
I am reading Sternberg's "Group Theory and Physics."
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7This is a consequence of Lagrange's theorem. – Tobias Kildetoft Jan 22 '16 at 07:13
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Thank you so much! Just looked at wikipedia article on Lagrange's theorem and there was the answer. Now, I know what cyclic subgroup is. – Youngsub Yoon Jan 22 '16 at 07:20
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Yeah, I took a quick look at the book you are reading, and it seems like it skips several important things in order to quickly get to those parts of interest to physics. – Tobias Kildetoft Jan 22 '16 at 07:22
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This doesn't directly address your question, but the notation $|G|$ is the standard notation for the number of elements in $G$, which we refer to as the order of $G$. I've actually never seen the notation $#G$. – manthanomen Jan 22 '16 at 07:27
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1@manthanomen It is the notation used in the book mentioned, and while not common, it does pop up here and there (more commonly for just sets than groups though). – Tobias Kildetoft Jan 22 '16 at 07:30
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I suggest you read the first chapter ot two of any introductory book on groups, especially on the topic of finite groups. – DanielWainfleet Jan 22 '16 at 08:40
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See also https://math.stackexchange.com/questions/28332/is-lagranges-theorem-the-most-basic-result-in-finite-group-theory – lhf Sep 07 '19 at 21:10