Exercise 10.2.3 (Finite subgroup of multiplicative group)

Question

Let $K$ be a field and let $H$ be a finite subgroup of $K^{	imes}$ of order $n$.
Prove that $H \subseteq U_n(K)$.

Hassaan Naeem · Accepted Answer

extbf{Solution:}
By 	extbf{Proposition 10.2.1} $H$ is cyclic, since it is finite. We can trivially see that $H$ is then isomorphic to $\mathbb{Z}/n\mathbb{Z}$ where $n$
is the order of $H$. 
Hence  We also have by 	extbf{Example 10.2.2} that
$U_n(K) = \{\alpha \in  K: \alpha^n= 1\}$, and it is also cyclic.

Richard Ganaye · Answer

\begin{proof} 
Let $x \in H$. Then  $\langle x angle = \{e,x,x^2,\ldots,x^{d-1}\}$, where $x^d = 1$, is a subgroup of $H$. By Lagrange Theorem, the cardinality $d = |\langle x angle|$ divides $n = |H|$. Since $x^d =1$ and $d \mid n$, then $n = k d$ for some integer $k$. We obtain $x^n = (x^d)^k = 1$. This shows that $x \in U_n(K) = \{\alpha \in K \mid \alpha^n = 1\}$. We have proved $H \subseteq  U_n(K)$.
\end{proof}

\medskip
Note: Since $H$ is a finite subgroup of $K^	imes$, $H$ is cyclic, so $H = \langle y angle$, where $y \in H$ has order $n$. This is not used in the previous proof.

Exercise 10.2.3 (Finite subgroup of multiplicative group)

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Exercise 10.2.3 (Finite subgroup of multiplicative group)

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