The form of the subgroup of a finite abelian $p$-group Let $G\cong \Bbb{Z}_{p^{r_1}}\oplus\Bbb{Z}_{p^{r_2}}\oplus\cdots \oplus \Bbb{Z}_{p^{r_s}}$ be a finite abelian $p$-group, where $r_1\geq r_2\geq \cdots \geq r_s\geq 1$.
Let $H\cong \Bbb{Z}_{p^{t_1}}\oplus\Bbb{Z}_{p^{t_2}}\oplus\cdots \oplus \Bbb{Z}_{p^{t_u}}$ be a subgroup of $G$,
where $t_1\geq t_2\geq \cdots \geq t_u\geq 1$.
Prove that $s\geq u$ and $r_i\geq t_i$ for each $i=1, 2, ..., u$.
It seems obviously. But I just can't prove it.
Thanks for any help.
 A: By Lemma II.2.5 in Hungerford's Algebra,
$$H[p]=\overbrace{\Bbb{Z}_p\oplus \cdots \oplus \Bbb{Z}_p}^{u\text{ times}}$$
is a subgroup of
$$G[p]=\overbrace{\Bbb{Z}_p\oplus \cdots \oplus \Bbb{Z}_p}^{s\text{ times}}$$
Note that both are elementary abelian group.
View them as $\Bbb{Z}$-module.
Since $G[p]$ and $H[p]$ are annihilated by the ideal $\langle p\rangle$ in $\Bbb{Z}$,
we can view $G[p]$ and $H[p]$ as vector space over $\Bbb{Z}/\langle p\rangle\cong \Bbb{Z}_p$.
(See Example (5) in page 338 in Dummit and Foote's Abstract Algebra.)
Hence, $H[p]$ is a subspace of $G[p]$ over $\Bbb{Z}_p$
and $u=\dim_{\Bbb{Z}_p}H[p]\leq \dim_{\Bbb{Z}_p}G[p]=s$.
Now, we can write $H\cong \Bbb{Z}_{p^{t_1}}\oplus\Bbb{Z}_{p^{t_2}}\oplus\cdots \oplus \Bbb{Z}_{p^{t_u}}\oplus \Bbb{Z}_{p^{t_{u+1}}}\oplus \cdots \oplus \Bbb{Z}_{p^{t_s}}$,
where $t_{u+1}=\cdots =t_s=0$.
Suppose that $r_j<t_j$ for some $j\in \{1, 2, ..., s\}$.
In this case,
by Lemma II.2.5 again,
$$p^{r_j}H\cong \Bbb{Z}_{p^{t_1-r_j}}\oplus\Bbb{Z}_{p^{t_2-r_j}}\oplus\cdots \oplus \Bbb{Z}_{p^{t_j-r_j}}$$
is a subgroup of
$$p^{r_j}G\cong \Bbb{Z}_{p^{r_1-r_j}}\oplus\Bbb{Z}_{p^{r_2-r_j}}\oplus\cdots \oplus \Bbb{Z}_{p^{r_{j-1}-r_j}}.$$
Which is impossible because the length of $p^{r_j}H$ is greater than $p^{r_j}G$.
A: Hint: Count how many elements of order $p^d$ there are in each of $G$ and $H$, for each $d\in\mathbb{N}$.  Since $H$ is a subgroup of $G$, the count for $H$ must be less than or equal to the count for $G$.
