finite subgroup

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

$xy\in K\mathit{\hspace{1em}}\text{for all}\mathit{\hspace{1em}}x,y\in K.$(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

$$b^r=b^s,r>s+1.$$

By (1),

$$K\ni b^{r-s-1}=b^{r-s}{b}^{-1}=e{b}^{-1}=b^{-1}.$$

Thus also  $a{b}^{-1}\in K$,  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 $G$.