Return to Theorems, Glossary, Homework Problems.

**Statement:**

If $m$ divides the order of a finite abelian group G, then G has a subgroup of order $m$.

**Proof:**

By Theorem 11.12, we can thing of G as being

(1)where not all primes $p_{i}$ need be distinct. Since $(p_{1})^{r_{1}} (p_{2})^{r_{2}} ... (p_{n})^{r_{n}}$ is the order of G, then $m$ must be of the form $(p_{1})^{s_{1}} (p_{2})^{s_{2}} ... (p_{n})^{r_{s}}$, where $0\leq s_{i} \leq r_{i}$. By Theorem 6.14, $(p_{i})^{r_{i}-s_{i}}$ generates a cyclic subgroup of $\mathbb{Z}_{(p_{i})^{r_{i}}}$ of order equal to the quotient of $(p_{i})^{r_{i}}$ by the gcd of $(p_{i})^{r_{i}}$ and $(p_{i})^{r_{i}-s_{i}}$. But the gcd of $(p_{i})^{r_{i}}$ and $(p_{i})^{r_{i}-s_{i}}$ is $(p_{i})^{r_{i}-s_{i}}$. Thus $(p_{i})^{r_{i}-s_{i}}$ generates a cyclic subgroup of $\mathbb{Z}_{(p_{i})^{r_{i}}}$ of order

(2)Recalling that $\langle a \rangle$ denotes the cyclic subgroup generated by a, we see that

(3)is required subgroup of order $m$.