Existence of subgroup of order power of prime in a finite abelian group?

Question

Say I have a finite abelian group $G$ such that $\left | G \right |=p_1^{n_1}...p_m^{n_m}$ where the $p_i$'s are distinct. Can I say that $G$ must have a subgroup of order $p_1^{n_1}$? I'm thinking that I can use the structure theorem to write $G$ as $\mathbb{Z}/p_1^{n_{1_{1}}}\times ...\times \mathbb{Z}/p_1^{n_{1_{s_{1}}}}\times...\times\mathbb{Z}/p_m^{n_{m_{1}}}\times ...\times \mathbb{Z}/p_m^{n_{m_{s_{m}}}}$ then just take $\mathbb{Z}/p_1^{n_{1_{1}}}\times ...\times \mathbb{Z}/p_1^{n_{1_{s_{1}}}}\times\left \{ 1 \right \}...\times\left \{ 1 \right \}$ where $\sum_i n_{1_{i}}=n_1$.

Is my reasoning correct? I would like to avoid using Sylow theorems.

Thanks in advance.

Answer 1

Some highlights:

First, you can take $\;K:=\Bbb Z/p_1^{n_1}\Bbb Z\times\{1\}\times\ldots\times\{1|]\;$ to get a subgroup of order $\;p_1^{n_1}\;$ , and now you can generalize this to each prime $\;p_1,..,p_m\;$ and their powers.

Next, use the basic lemma that says that a finite $\;p\,-$ group of order $\;p^n\;$ has a (normal if you will, even when we're not in the abelian case!) subgroup of order $\;p^k\;$ , for any $\;0\le k\le n\;$ .

Finally, just use the direct product decomposition to get a subgroup of any order dividing the group's.

Answer 2

Here is a roadmap.

Let $p$ be a prime dividing the order of $G$.

• Let $P = \{ g \in G : ord(g) \text{ is a power of $p$} \}$.

• Then $P$ is a subgroup of $G$ (because $G$ is abelian).

• The order of $P$ is a power of $p$ (by Cauchy's theorem).

• The order of $P$ is the largest power of $p$ that divides the order of $G$ (by Cauchy's theorem applied to $G/P$).