1

I am reading I.N.Herstein-Topics in Algebra

Question:

For $G$ a finite group. Let $|G|=n$, then $O(a)|n \enspace \forall a\in G$.

Note:
I have already proved that if $G$ is a finite group, then $\exists N\in \mathbb{N}$ such that $a^N=e \enspace \forall a \in G$ and $N\leq n$.

Proof:
Suppose $\exists a\in G$ and a minimal $m\in \mathbb{N}$ such that $a^m=e$ and $m\nmid n$.

Given $G$ is finite. Therefore, we have $$a^n=e\\\text{By using division algorthim,}\enspace n=km+r\enspace\text {for some k}\in \mathbb{N} \\ \implies a^{n-km}.\{a^m\}^k=e \\ \implies a^{n-km}.e=e \\ \implies a^{n-km}=e$$ This Contradicts minimality of $m$.

Therefore, $O(a)|n \enspace \forall a\in G$.
This also proves that $N|n$.

Is My proof Correct? Any Improvements and suggestions are appreciated.

Kumar
  • 1,167
  • 1
    Sorry, but how does the note you wrote prove that $a^n=e$? The note says you only know that there is $N\leq n$ such that $a^N=e$. It doesn't simply imply that $a^n=e$. – Mark Jul 20 '19 at 14:01
  • @Mark Thanks for the correction. I got it. $a^n=e$ is not implied by the fact that $a^N=e$ but it is implied by the finiteness of $G$. – Kumar Jul 20 '19 at 14:06
  • 1
    It is true that if $G$ is finite then $a^{|G|}=e$ but the proof is not so trivial. Do you know about Lagrange's theorem? That if $H$ is a subgroup then $|H|$ divides $|G|$? Everything follows from this, including what you need to prove. – Mark Jul 20 '19 at 14:08
  • 1
    See also https://math.stackexchange.com/questions/28332/is-lagranges-theorem-the-most-basic-result-in-finite-group-theory – lhf Jul 20 '19 at 14:13
  • @Mark I Know Lagrange's Theorem. But I haven't proved it yet. and Herstein asks to prove the above fact without the use of Lagrange's Theorem. – Kumar Jul 20 '19 at 14:14
  • Ok, I'm just confused how do you know that $a^n=e$ then. The proof I know actually follows from the fact that the order of $a$ must divide $n$ (if $n=O(a)\times d$ then obviously $a^n=(a^{O(a)})^d=e^d=e$), which is what you need to prove. Maybe there are other proofs, I don't know. – Mark Jul 20 '19 at 14:17
  • @Mark I have concluded by the help of examples. But I think I have overkilled it there itself. – Kumar Jul 20 '19 at 14:21
  • 1
    You cannot start with $a^n=e$. – Wuestenfux Jul 20 '19 at 14:24
  • Well, examples are not a proof. To be honest I have never seen a proof of the statement you want to prove without Lagrange's theorem. – Mark Jul 20 '19 at 14:24

0 Answers0