Theorem 11.16

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)
\begin{align} \mathbb{Z}_{(p_1)^{r_1}} \times \mathbb{Z}_{(p_2)^{r_2}} \times...\times \mathbb{Z}_{(p_n)^{r_n}} \end{align}

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)
$$[(p_{i})^{r_{i}}]/[(p_{i})^{r_{i}-s_{i}}]=(p_{i})^{s_i}$$

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

(3)
\begin{align} \langle (p_{1})^{r_{1}-s_{1}} \rangle \times \langle (p_{2})^{r_{2}-s_{2}} \rangle \times...\times \langle (p_{i})^{r_{i}-s_{i}} \rangle \end{align}

is required subgroup of order $m$.