# finite subgroup

## Primary tabs

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\textbf{Theorem.}\, A non-empty finite subset $K$ of a group $G$ is a subgroup of $G$ if and only if
\begin{align}
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\emph{Proof.}\, The condition (1) is apparently true if $K$ is a subgroup.\, Conversely, suppose that a nonempty finite subset $K$ of the group $G$ satisfies (1).\, Let $a$ and $b$ be arbitrary elements of $K$.\, By (1), all (\PMlinkescapetext{positive}) powers of $b$ belong to $K$.\, Because of the finiteness of $K$, there exist positive integers $r,\,s$ such that
$$b^r \;=\; b^s, \quad r \;>\; s\!+\!1.$$
By (1),
$$K \;\ni\; b^{r-s-1} \;=\; b^{r-s}b^{-1} \;=\; eb^{-1} \;=\; b^{-1}.$$
Thus also\, $ab^{-1} \in K$,\, whence, by the theorem of the \PMlinkid{parent entry}{1045}, $K$ is a subgroup of $G$.\\

\textbf{Example.}\, The multiplicative group $G$ of all nonzero complex numbers has the finite multiplicative subset
\,$\{1,\,-1,\,i,\,-i\}$,\, which has to be a subgroup of $G$.
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