A result in Group Theory says that every group of prime order is cyclic. I understand the proof on: http://planetmath.org/proofthateverygroupofprimeorderiscyclic but i dont understand why the order of the element must exist. example:Consider G={e,a,b}, $a\cdot b=e$ ,$b \cdot a=e$,$a \cdot a=a$,$b \cdot b=b$,where e is the identity element.clearly G is of prime order( 3) but i dont see how the group is cyclic
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1What you've written down isn't a group of order 3: if $aa=a$ then $a=e$ – Matthew Towers Jun 08 '18 at 19:21
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Duplicate for the title question, but I guess the work here was rather clearing up the user's mistake rather than answering the question. – rschwieb Jun 08 '18 at 20:02
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Does this answer your question? Proof that all abelian simple groups are cyclic groups of prime order – tryst with freedom Mar 10 '22 at 17:14
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@Buraian No, the answer to that question was not what I wanted to clarify. The first answer below is what I was unsure of. – kishan17 Mar 11 '22 at 15:05
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To see that the order of an element in a finite group exists, let $ G $ be a finite group and $ a $ an arbitrary non-identity element in that group. Since $ G $ is finite, the sequence $ a, a^2, a^3, \dots $ must have repeats. Let $ m $ be minimal such that $ a^m = a^n $ for some $ n < m $. Then $ m - n > 0 $ and $ a^{m - n} = 1 $. (In particular, you have $ a = 1 $ in your example, hence why it doesn't work).
Tom Miller
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Your example is inconsistent: you can't have $a^2=a$ for any non-identity element of a group because the cancellation property implies that $a=e$. Indeed, even by your rules $ e = ab = (a^2)b = a(ab) = a$.
That the order of an element divides the order of the group is Lagrange's theorem.
Chappers
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