# finite subgroup

Theorem.  A non-empty finite subset $K$ of a group $G$ is a subgroup of $G$ if and only if

 $\displaystyle xy\in K\quad\mbox{for all}\quad x,\,y\in K.$ (1)

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 () 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 http://planetmath.org/node/1045parent entry, $K$ is a subgroup of $G$.

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 $G$.

Title finite subgroup FiniteSubgroup 2013-03-22 18:57:02 2013-03-22 18:57:02 pahio (2872) pahio (2872) 5 pahio (2872) Theorem msc 20A05 criterion for finite subgroup finite subgroup criterion