# 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.

• I think you can get the $s\geqslant u$ by considering the dimension of the natural $\mathbb{F}_p$ vector spaces $V:=\{x\in G|px=0\}$, $U:=\{x\in H|px=0\}$. And I think that passing to $pG$, $pH$ may help with the indices. Commented Mar 5, 2017 at 17:31
• @ancientmathematician did you mean that use induction on $n$ for the second part, where $|G|=p^n$? Commented Mar 5, 2017 at 18:42
• @bhafa: induction of $|G|$. Commented Mar 6, 2017 at 7:26

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 $p^{r_j}G$.

• As the proof (i)$\Rightarrow$(iv) of Theorem II.1.1 in Hungerford's Algebra, we have a commutative diagram \begin{align} X & \stackrel{\iota}{\rightarrow} & F \\ & \searrow & \downarrow \overline{f}\\ & & G \end{align} where $F$ is a free abelian group of rank $s$. Consider $\overline{f}^{-1}(H)\leq F$. By Theorem II.1.6, $\overline{f}^{-1}(H)$ is also a free abelian group of rank $u\leq s$. Commented Mar 15, 2017 at 3:54
• Suppose that $\{y_1, y_2, ..., y_u\}$ is a basis of $\overline{f}^{-1}(H)$. By Theorem II.1.6 again, there is a basis $\{d_1 y_1, d_2 y_2, ..., d_w y_w\}$ for the kernel of the homomorphism $\overline{f}|_{\overline{f}^{-1}(H)}$, where $w\leq u$. Commented Mar 15, 2017 at 3:55
• By First Isomorphism Theorem, \begin{eqnarray*} H &\stackrel{\overline{f}\text{ is onto}}{=}& \overline{f}(\overline{f}^{-1}(H)) =\text{Im }\overline{f}|_{\overline{f}^{-1}(H)} \cong \overline{f}^{-1}(H)/\ker{\overline{f}|_{\overline{f}^{-1}(H)}}\\ &\cong& \frac{\overbrace{y_1 \Bbb{Z}\oplus y_2 \Bbb{Z}\oplus \cdots \oplus y_u \Bbb{Z}}^{u \text{ times}}}{\underbrace{d_1 y_1 \Bbb{Z}\oplus d_2 y_2 \Bbb{Z}\oplus \cdots \oplus d_w y_w\Bbb{Z}}_{w \text{ times}}} \cong \Bbb{Z}_{d_1}\oplus \Bbb{Z}_{d_2}\oplus \cdots \oplus \Bbb{Z}_{d_w}\oplus \Bbb{Z} \oplus \cdots, \end{eqnarray*} Commented Mar 15, 2017 at 3:56
• where $d_1\mid d_2\mid \cdots \mid d_w$. Since $H$ is a subgroup of a finite group $G$, it must be $w=u$. Since $H$ is a subgroup of a $p$-group $G$, it must be $d_1=p^{t_1}, d_2=p^{t_2}, ..., d_w=d_u=p^{t_u}$. Commented Mar 15, 2017 at 3:57

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 $G$.

• Could you please give me a little more hint or reference? Commented Mar 6, 2017 at 16:34
• I think that $d$ can't be arbitrary. For example, $H\cong \Bbb{Z}_4\oplus \Bbb{Z}_4\leq \Bbb{Z}_8\oplus \Bbb{Z}_2$. I think that your idea is suppose that there is $r_j<t_j$ for some $j$. Then the number of elements of order $p^d$ in $H$ is greater than in $G$ for some $d$. Commented Mar 7, 2017 at 18:02