## Properties of the image and kernel of the p-power map on a finite abelian group

Let $A$ be a finite abelian group (written multiplicatively) and let $p$ be a prime. Let $A^p = \{ a^p \ |\ a \in A \}$ and $A_p = \{ x \in A \ |\ x^p = 1 \}$. (I.e. $A^p$ and $A_p$ are the image and kernel of the $p$-power map, respectively.

1. Prove that $A/A^p \cong A_p$. (Show that they are both elementary abelian and have the same order.)
2. Prove that the number of subgroups of $A$ of order $p$ equals the number of subgroups of index $p$. (Reduce to the case where $A$ is an elementary abelian $p$-group.)

1. We begin with some lemmas.

Lemma 1: Let $A$ and $B$ be abelian groups and let $\theta : A \rightarrow B$ be an isomorphism. Then there exists an isomorphism $\theta^\prime : A/A^p \rightarrow B/B^p$. Proof: Let $\pi$ denote the natural projection from $Z$ to $Z/Z^p$ (for $Z \in \{A,B\}$). If $x \in A^p$, then $x = y^p$ for some $y \in A$. Then $(\pi \circ \theta)(x) = (\pi \circ \theta)(y^p)$ $= \pi(\theta(y)^p)$ $= B^p$, so that $A^p \leq \mathsf{ker}\ (\pi \circ \theta)$. By the remarks on page 100 in D&F, there exists a unique group homomorphism $\theta^\prime : A/A^p \rightarrow B/B^p$ such that $\theta^\prime \circ \pi = \pi \circ \theta$. ($\theta^\prime$ is injective) Suppose $xA^p \in \mathsf{ker}\ \theta^\prime$. Then $\theta^\prime(\pi(x)) = 1$, thus $(\theta^\prime \circ \pi)(x) = 1$, so that $(\pi \circ \theta)(x) = 1$, hence $\theta(x) \in B^p$. Now since $\theta$ is surjective, $\theta(x) = \theta(y)^p$ for some $y \in A$. Thus $x = y^p$, so that $x \in A^p$. Thus $\mathsf{ker}\ \theta^\prime = 1$, so that $\theta^\prime$ is injective. ($\theta^\prime$ is surjective) Let $xB^p \in B/B^p$. Now $xB^p = (\pi \circ \theta)(y)$ for some $y \in A$, so that $xB^p = \theta^\prime(\pi(y))$. Hence $\theta^\prime$ is surjective. $\square$

Lemma 2: Let $A$ and $B$ be abelian groups and let $\theta : A \rightarrow B$ be an isomorphism. Then $\theta[A_p] = B_p$. Proof: $(\subseteq)$ If $x \in \theta[A_p]$, then $x = \theta(y)$ where $y \in A_p$. Now $y^p = 1$, so that $\theta(y)^p = 1$, hence $x^p = 1$. Thus $x \in B_p$. $(\supseteq)$ Let $x \in B_p$. Then x^p = 1\$. Since $\theta$ is surjective, $x = \theta(y)$ for some $y \in A$. Now $\theta(y^p) = \theta(y)^p = x^p = 1$, and since $\theta$ is injective, $y^p = 1$. Thus $y \in A_p$, hence $x \in \theta[A_p]$. $\square$

Lemma 3: Let $G$ be a finite abelian group of order $n$, let $m$ be an integer, and let $\varphi : G \rightarrow G$ denote the $m$-power homomorphism. If $\mathsf{gcd}(m,n) = 1$, then $\varphi$ is an isomorphism. Proof: Since $G$ is finite, it suffices to show that $\varphi$ is injective. Let $x \in \mathsf{ker}\ \varphi$. Then $x^m = 1$, so that $|x|$ divides $m$. By Lagrange, $|x|$ divides $n$. Thus $|x| = 1$, hence $x = 1$, and $\varphi$ is injective. $\square$

Now to the main result. By Theorem 5 in the text, $A \cong A_1 \times A_2$ where $A_1$ is a $p$-group of rank $t$ and $p$ does not divide $|A_2|$. By a lemma to the previous exercise, $\varphi = \varphi_1 \times \varphi_2$, where each map is the $p$-power map on $A_1 \times A_2$, $A_1$, and $A_2$, respectively.

We have $\mathsf{ker}\ \varphi = \mathsf{ker}(\varphi_1 \times \varphi_2)$ $= (\mathsf{ker}\ \varphi_1) \times (\mathsf{ker}\ \varphi_2)$ $\cong Z_{p^t} \times 1$ $\cong Z_{p^t}$ by the previous exercise.

Similarly, $(A_1 \times A_2)/\mathsf{im}\ \varphi = (A_1 \times A_2)/(\mathsf{im}\ \varphi_1 \times \mathsf{im}\ \varphi_2)$ $\cong (A_1/\mathsf{im}\ \varphi_1) \times (A_2/\mathsf{im}\ \varphi_2)$ $\cong Z_{p^t} \times 1$ $\cong Z_{p^t}$, again using the previous exercise.

Now the first conclusion follows using Lemmas 1 and 2.

2. We begin with some more lemmas.

Lemma 4: Let $E_{p^k}$ be the elementary abelian group of order $p^k$, and suppose $E_{p^k} = \langle S \rangle$ where $|S| = k$. Finally let $G$ be a group. If $\overline{\varphi} : S \rightarrow G$ is a mapping such that $|\overline{\varphi}(s)| = p$ for each $s \in S$ and $\overline{\varphi}(s) \overline{\varphi}(t) = \overline{\varphi}(t) \overline{\varphi}(s)$ for all $s,t \in S$, then $\overline{\varphi}$ extends uniquely to a group homomorphism $\varphi : E_{p^k} \rightarrow G$. Proof: Every element of $E_{p^k}$ can be written uniquely as $x = s_1^{a_1}s_2^{a_2} \cdots s_k^{a_k}$, where $s_1 \in S$ and $0 \leq a_k < p$. Define $\varphi(x) = \overline{\varphi}(s_1)^{a_1}\overline{\varphi}(s_2)^{a_2} \cdots \overline{\varphi}(s_k)^{a_k}$. $\varphi$ is well defined since the $S$-expansion of $x$ is unique, and is a homomorphism since the $\overline{\varphi}(s)$ commute with one another. To see uniqueness, suppose $\psi : E_{p^k} \rightarrow G$ is a group homomorphism such that $\psi(s) = \varphi(s)$ for all $s \in S$. Then $\psi(x) = \psi(s_1^{a_1}s_2^{a_2} \cdots s_k^{a_k})$ $= \psi(s_1)^{a_1} \psi(s_2)^{a_2} \cdots \psi(s_k)^{a_k}$ $= \overline{\varphi}(s_1)^{a_1}\overline{\varphi}(s_2)^{a_2} \cdots \overline{\varphi}(s_k)^{a_k}$ $= \varphi(x)$. Thus $\varphi$ is unique. $\square$

Note that every subgroup of $A$ of order $p$ is contained in $\mathsf{ker}\ \varphi$, so that in fact the order $p$ subgroups of $A$ and $\mathsf{ker}\ \varphi$ coincide. Now because $\mathsf{ker}\ \varphi \cong Z_{p^t}$ is elementary abelian, every nonidentity element has order $p$ and thus generates an order $p$ subgroup. Each such subgroup is generated by $p-1$ elements; thus there are $(p^t-1)/(p-1)$ order $p$ subgroups in $\mathsf{ker}\ \varphi$, thus in $A$.

Let $B \leq A$ be a subgroup of index $p$. Now let $x \in A^p$, say $x = y^p$, and suppose $x \notin B$. Then $y \notin B$, so that $A = \langle y \rangle B$. Now $A/B$ is a group of order $p$, so that $(yB)^p = xB = B$; then $x \in B$, a contradiction. Thus $A^p \leq B$. Moreover, by the Third Isomorphism Theorem, $[A/A^p : B/A^p] = [A:B] = p$. By the Lattice Isomorphism Theorem, the index $p$ subgroups of $A/A^p \cong Z_p^t$ correspond precisely to the index $p$ subgroups of $A$. In particular, it suffices to count the order and index $p$ subgroups in an elementary abelian $p$-group.

Now $\mathsf{Aut}(Z_p^t)$ acts on $\mathcal{M} = \{ M \leq Z_p^t \ |\ [Z_p^t : M] = p \}$ by $\varphi \cdot M = \varphi[M]$. We claim that this action is transitive. To that end, let $M_1, M_2 \leq Z_p^t$ be index $p$ subgroups. Now $M_1 \cong M_2 \cong Z_p^{t-1}$ since $M_1$ and $M_2$ are elementary abelian, thus $M_1 = \langle S_1 \rangle$ and $M_2 = \langle S_2 \rangle$ where $|S_1| = |S_2| = t-1$. Let $\overline{\psi} : S_1 \rightarrow S_2$ be a bijection, and choose some $t_1 \in Z_p^t \setminus M_1$ and $t_2 \in Z_p^t \setminus M_2$. Now $\overline{\psi} \cup \{(t_1,t_2)\}$ extends by Lemma 4 to a homomorphism $\psi : Z_p^t \rightarrow Z_p^t$. Clearly $\psi$ is surjective, hence an isomorphism, so that $\psi \in \mathsf{Aut}(Z_p^t)$. Moreover, we see that $\psi[M_1] = M_2$. Thus this action is transitive, and we have $\mathsf{Aut}(Z_p^t) \cdot M = \mathcal{M}$ for all index $p$ subgroups $M$.

Now let $M \leq Z_p^t$ be an arbitrary index $p$ subgroup and let $\varphi$ be an automorphism of $Z_p^t$ which stabilizes $M$. As above, $M = \langle S \rangle$ for some set $S$ with $|S| = t-1$. Moreover, if $x \in Z_p^t \setminus M$ then $Z_p^t = \langle S \cup \{x\} \rangle$ since $M$ is maximal. By Lemma 4, $\varphi$ is determined uniquely by is action on $S$ and $x$. We see also that since $\varphi$ stabilizes $M$, we must have $\varphi(s) \in M$ for each $s \in S$. Since $M \cong Z_p^{t-1}$ and $\varphi|_M \in \mathsf{Aut}(M)$, there are $\prod_{i=0}^{t-2}(p^{t-1} - p^{t-2-i})$ distinct choices we can make for $\varphi[S]$. Moreover, $\varphi(x)$ may be chosen arbitrarily so long as $\varphi(x) \notin M$; thus there are $p^t - p^{t-1} = p^{t-1}(p-1)$ choices for $\varphi(x)$. Thus in total $\mathsf{Stab}(M)$ contains $(p-1)p^{t-1} \prod_{i=0}^{t-2}(p^{t-1} - p^{t-2-i})$ elements.

By the Orbit-Stabilizer Theorem and Lagrange's Theorem, $|\mathcal{M}| = (\prod_{i=0}^{t-1} (p^t - p^{t-1-i}))/((p-1)p^{t-1} \prod_{i=0}^{t-2}(p^{t-1} - p^{t-2-i}))$ $= (\prod_{i=1}^t (p^t - p^{t-i}))/((p-1) \prod_{i=1}^{t-1}(p^{t} - p^{t-i}))$ $= (p^t-1)/(p-1)$.

Thus the theorem is proved.

• quadro  On March 8, 2011 at 2:37 am

in the last parag.,How does the last equality follow? where does p^(t-1) factor go?

• nbloomf  On March 8, 2011 at 7:59 am

Good question. I’ll try to find the problem later today.

• quadro  On March 9, 2011 at 4:07 am

Oh sorry i see my misteke there is no problem,everything is ok.

• nbloomf  On March 9, 2011 at 10:42 am

Maybe you could explain it to me.🙂 I am still convinced there is a mistake.

• quadro  On March 14, 2011 at 6:31 am

i mean the last calculations. you have already modified them now its true i guess.
by the way in some solutions to the problems as in this one you say “By the remarks on page … ” what does “page …” refer to?how can i get to these pages?

• nbloomf  On March 14, 2011 at 9:19 am

Sorry- the page references are to Dummit and Foote’s “Abstract Algebra”, 3rd edition.

I did fix a mistake in the expressions which count the automorphisms, but unless I am missing something I think it is still wrong. The strategy seems to be okay, but that pesky $p^{t-1}$ factor doesn’t go away. Either I’m making an arithmetic error, or something is counted wrong.

• quadro  On March 14, 2011 at 4:34 pm

in the paragraph just before the last one, the product over i should start from i=0 to i=t-2 as stated in 4 lines above that line.
nevertheless in the last paragraph you start the product from i=0 but this time the general term over that product should be “p^(t-1)-p^(t-2-i)” not “p^(t-1)-p^(t-1-i)”.this solves the problem i guess now.
by the way i wonder where did you find such an important theorem?is this from a special book or an article or?

• nbloomf  On March 15, 2011 at 8:32 am

Thanks! All of these problems come from Dummit and Foote’s “Abstract Algebra”, 3rd edition. Eventually I will start pulling problems from other books too.

During the course of trying to solve this problem, I found out that in fact, in an elementary abelian group of order $p^t$, the number of subgroups of order $p^k$ and $p^{t-k}$ is the same for all $k$. Moreover, in a certain sense, this result is a generalization of the binomial theorem. The Wiki page on q-analogs gives more information on this.

• Alex Youcis  On August 25, 2011 at 12:44 am

Hey, I just thought I’d let you know. There is a short proof for the number of $p$-indexed subsets of the elementary abelian group of order $p^n$ assuming you know the rudiments of linear algebra. Namely, if $A$ is an elementary abelian group of order $p^n$ then it’s easy to see that $A$ is (canonically) a vector space over $\mathbb{F}_p$ and then note that subgroups and subspaces coincide (since multiplication corresponds to addition). In particular, the subgroups of $A$ of order $p$ are precisely the subspaces of $A$ of dimension $1$ and those of index $p$ are those of order $n-1$. But, it is a simple fact of linear algebra that if $V$ is an $n$-dimensional $\mathbb{F}_q$-space then the number of $k$-dimensional subspaces is equal to $\displaystyle \frac{(q^n-1)(q^n-q)\cdots(q^n-q^{k-1})}{(q^{k}-1)(q^{k}-q)\cdots (q^k-q^{k-1})}$–it’s a simple counting argument about choosing $k$ linearly independent subsets (it’s on pg. 412 of Dummit and Foote).

• Alex Youcis  On August 25, 2011 at 12:47 am

And of course, plugging in $q=p$ and $k=1,n-1$ gives the desired result (not only that their equal, but that they are equal to $(p^n-1)(p-1)^{-1}$.

• nbloomf  On August 25, 2011 at 8:37 am

Thanks!