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

for all

(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 (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 parent 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 $G$ .