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I am well aware there are many proofs out there which address this result. However, my question is regarding the subtlety of one of the proofs. The proof of interest is the following:

Let $G$ be any finite multiplicative subgroup of field $F$. Then let $\epsilon$ be the exponent of $G$. For any $x \in G$, we have $x ^ \epsilon = 1$. Since this equation has at most $\epsilon$ roots, $G$ has at most $\epsilon$ elements (i.e. $|G|≤\epsilon$). And for any finite group, $\epsilon≤|G|$. Therefore, $|G|=\epsilon$. We then use the known result that any finite abelian group ($G$) will contain an element of order $\epsilon$ to arrive at our final result.

My question is: why are we able to deduce that there are at most $\epsilon$ roots in the equation? The counter-example I can think of is the Klein-4 group where there are 3 roots with order 2.

sunnydk
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    Because you are in a field, and the solutions are all roots of the polynomial $x^{\epsilon}-1$. Polynomials in a field cannot have more solutions than the degree. – Arturo Magidin Jun 28 '23 at 02:33
  • Pardon my ignorance, how does being in a field apply to the fact that the solutions are all roots of a polynomial? I do realise I am missing this link, but I cannot think of why this link is as such! – sunnydk Jun 28 '23 at 02:34
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    The elements of $G$ are elements of the field. The multiplicative identity of the field is the identity of $G$. When you say $G$ has "exponent $n$", you are saying each element $a$ of $G$ satisfies $a^n=1$. That means they are all roots of $x^n-1$, viewed as a polynomial over the field. What you are missing is that $G$ is not just some random group, it is a subgroup of the multiplicative group of the field. – Arturo Magidin Jun 28 '23 at 02:37
  • ah i see, so you can only view $a$ as solutions to a polynomial when $a$ is an element of a field, not just a group. But just to clarify, is this because polynomials can only be over a ring or field? – sunnydk Jun 28 '23 at 02:48
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    Polynomial functions require two operations to make sense. – Arturo Magidin Jun 28 '23 at 02:50
  • @ArturoMagidin Thank you! – sunnydk Jun 28 '23 at 02:52
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    Since your question has been answered, I will close as a duplicate. – Arturo Magidin Jun 28 '23 at 03:52

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