Problem 5.2.8 (The number of subgroups of $A$ of order $p$ equals the numbers of subgroups of $A$ of index $p$ ($A$ abelian))

Proof. Let $A$ be a finite abelian group and let $p$ be a prime.

(a)

By Theorem 5, if $|A|=p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}\cdots p_k^{{\alpha }_k}$, then $$\begin{array}{lll}\hfill & A\simeq X_1\times X_2\times \cdots X_k,& \hfill \text{(1)}\end{array}$$

where $|X_i|=p_i^{{\alpha }_i}$, such that for each $X_i$,

$$\begin{array}{lll}\hfill & X_i\simeq Z_{p_i^{{\beta }_1}}\times Z_{p_i^{{\beta }_2}}\times \cdots \times Z_{p_i^{{\beta }_t}},& \hfill \text{(2)}\end{array}$$

with ${\beta }_1\ge {\beta }_2\ge \cdots \ge {\beta }_t\ge 1$ and ${\beta }_1+{\beta }_2+\cdots +{\beta }_t={\alpha }_i$ (where $t$ and ${\beta }_1,\dots ,{\beta }_t$ depend on $i$)

Consider the maps

$$\phi \{\begin{array}{ccc}\hfill A\hfill & \hfill \to \hfill & A\hfill \\ \hfill x\hfill & \hfill \mapsto \hfill & x^p\hfill \end{array},{\phi }_i\{\begin{array}{ccc}\hfill X_i\hfill & \hfill \to \hfill & X_i\hfill \\ \hfill x\hfill & \hfill \mapsto \hfill & x^p\hfill \end{array}(1\le i\le k).$$

Then, by (1)

$$\begin{array}{lll}\hfill A^p=\mathrm{im}(\phi )& =\mathrm{im}({\phi }_1)\times \mathrm{im}({\phi }_2)\times \cdots \times \mathrm{im}({\phi }_k),& \hfill \text{(3)}\\ \hfill A_p=\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )& =\ker <!--\; FUNCTION\; APPLICATION\; -->({\phi }_1)\times \ker <!--\; FUNCTION\; APPLICATION\; -->({\phi }_2)\times \cdots \times \ker <!--\; FUNCTION\; APPLICATION\; -->({\phi }_k),& \hfill \text{(4)}\end{array}$$

and by Exercise 5.1.14,

$$\begin{array}{lll}\hfill A∕A^p=A∕\mathrm{im}(\phi )\simeq X_1∕\mathrm{im}({\phi }_1)\times X_2∕\mathrm{im}({\phi }_2)\times \cdots \times X_k∕\mathrm{im}({\phi }_k)& & \hfill \text{(5)}\end{array}$$

• Suppose that $p\ne p_i$, for some fixed index $i$.

  For some fixed index $j$, put $q=p_i\ne p$ and $\beta ={\beta }_j$. Consider the restriction $\chi ={\chi }_{i,j}$ of ${\phi }_i$ given by

  $$\chi \{\begin{array}{ccc}\hfill Z_{q^{\beta }}\hfill & \hfill \to \hfill & Z_{q^{\beta }}\hfill \\ \hfill x\hfill & \hfill \mapsto \hfill & x^p\hfill \end{array}$$

  Write $Z_{q^{\beta }}=⟨y⟩$, where $|y|=q^{\beta }$. Let $x\in Z_{q^{\beta }}$, so that $x=y^k$ for some integer $k$.

  • If $x\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\chi )$, then $x^p=1$, thus $y^{\mathit{𝑝𝑘}}=1$. Since $|y|=q^{\beta }$, $q^{\beta }\mid \mathit{𝑝𝑘}$, where $p,q$ are distinct primes, thus $q^{\beta }\mid k$, so $x=1$. This shows $\ker <!--\; FUNCTION\; APPLICATION\; -->(\chi )=\{1\}$ and $\chi$ injective.
  • Since $Z_{q^{\beta }}$ is finite, $\chi$ is also surjective.

  Therefore $\chi$ is bijective, so

  $$\ker <!--\; FUNCTION\; APPLICATION\; -->(\chi )=\{1\},Z_{q^{\beta }}∕\mathrm{im}(\chi )=\{1\}.$$

  Then, by (2),

  $$\ker <!--\; FUNCTION\; APPLICATION\; -->({\phi }_i)=\{1\},X_i∕\mathrm{im}({\phi }_i)=\{1\}.$$

• Suppose now $p=p_i$. By Exercise 7,

  $$\begin{array}{lll}\hfill \ker <!--\; FUNCTION\; APPLICATION\; -->({\phi }_i)\simeq {(Z_p)}^t\simeq X_i∕\mathrm{im}({\phi }_i).& & \hfill \text{(6)}\end{array}$$

  (where $t$ is the number of elementary divisors associate to the prime $p=p_i$).

Then all factors in (5) are trivial, except perhaps one if $p=p_i$ for some index $i$. Therefore (4) and (5) give

$$\begin{array}{llll}\hfill A∕A^p\simeq \{1\}\simeq A_p& \text{if }p\notin \{p_1,p_2,\dots ,p_k\},& \hfill & \\ \hfill A∕A^p\simeq {(Z_p)}^t\simeq A_p& \text{if }p=p_i\text{ for some }i\in [[1,k]]& \hfill & \end{array}$$

In both cases,

$$A∕A^p\simeq A_p$$

are isomorphic elementary abelian $p$-groups.

(b)

First we deal with the case where $A=Z_p^t$ is an elementary abelian group. In additive notations, $A\simeq {𝔽}_p^t$ (the additive group of the vector space $E={𝔽}_p^t$).

The subgroups of $A$ are the lines of $E$. There are $p^t-1$ nonzero vectors in $E$, and two such vectors $v,w$ generate the same line if there is $\lambda \in {𝔽}_p^{\ast }$ such that $w=\mathit{𝜆𝑣}$, so there are

$$N_p=\frac{p^t-1}{p-1}$$

subgroups of order $p$ in $A=Z_p^t$.

The subgroups $H$ of index $p$ are the subspaces $H$ such that $E∕H\simeq {𝔽}_p$ has dimension $1$, i.e., are the hyperplanes of $E$. Every hyperplane has an equation $a_1{x}_1+a_2{x}_2+\cdots +a_t{x}_t=0$, where $(a_1,a_2,\dots ,a_t)\in {𝔽}^t-\{(0,0,,\dots ,0)\}$, and two equations $a_1{x}_1+a_2{x}_2+\cdots +a_t{x}_t=0$ and $a_1^{\prime }{x}_1+a_2^{\prime }{x}_2+\cdots +a_t^{\prime }{x}_t=0$ represent the same hyperplane if and only if there is some $\lambda \in {𝔽}^{\ast }$ such that $a_i^{\prime }=\lambda a_i,i=1,2,\dots ,t$. Therefore the number $M_p$ of hyperplanes (or subgroups of index $p$) is

$$M_p=\frac{p^t-1}{p-1}.$$

Now we deal with the general case.

• If $p\notin \{p_1,p_2,\dots ,p_n\}$, then $A_p=\{1\}$, so that there is no element of order $p$. Moreover $A^p=A$. Supppose that $H$ is a subgroup of $A$ of index $p$, so that $|A:H|=p$. Then $p$ divides $|A|=p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}\cdots p_k^{{\alpha }_k}$, so $p\in \{p_1,\dots ,p_n\}$. This is a contradiction, which proves that there are no subgroups of index $p$. So the number of elements of $A$ of order $p$ equals $0$, which is also the number of subgroups of $A$ of index $p$.

• Suppose that $p\in \{p_1,p_2,\dots ,p_n\}$, so that $p=p_i$ for some index $i$. By part (a),

  $$A_p\simeq {(Z_p)}^t\simeq A∕A^p.$$

  A subgroup of $A$ of order $p$ is generated by some element $y\in A$ of order $p$. Then $y\ne 1$ and $y^p=1$, so $y\in A_p-\{1\}$. Two elements $y,y^{\prime }\in A_p-\{1\}$ generate the same subgroup if and only if $y^{\prime }=y^k$, where $k=1,2,\dots ,p-1$. So there are exactly

  $$N_p=\frac{p^t-1}{p-1}$$

  elements of order $p$ in $A$.

  Consider now a subgroup $H\le A$ of index $p$, so that $|A∕H|=p$. For every $a\in A$, ${(\mathit{𝑎𝐻})}^p=H$, thus $a^{p}H=H$ and $a^p\in H$. This is true for any element $a\in A$, this shows

  $$A^p\subseteq H.$$

  By the Third Isomorphism Theorem, there are as many subgroups of $A∕A^p$ of index $p$ than subgroups of $A$ of index $p$ which contain $A^p$. But every subgroup of $A$ of index $p$ contains $A^p$. Therefore there are as many subgroups of $A$ of index $p$ that subgroups of $A∕A^p\simeq {(Z_p)}^t$ of index $p$, that is

  $$M_p=\frac{p^t-1}{p-1}$$

  subgroups of index $p$ in $A$.

  In conclusion, the number $N$ of subgroups of $A$ of order $p$ equals the numbers of subgroups of $A$ of index $p$, and

  $$N=\frac{p^t-1}{p-1},$$

  where $t$ is the number of elementary divisors associate to the prime $p=p_i$.