At page (24) of the book Finite Group Theory by I.Martin.Issacs, one finds the statement:

In the situation of Corollary 1.25, the number of subgroups of $G$ having order $p^b$ is congruent to 1 modulo $p$, where $|G|=p^am$ with $(p,m)=1$ and $b<a$.

I think the statements in the quotation and in the title are equivalent.
The author says that this result is not especially difficult to prove, thus he has decided not to put a proof there. But, after some attempts to no avail, I think I had better ask for some help here.
Thus far I have tried, following the tone adapted in the book, to define an action on the set of all subgroups of order $p^b$, and then use the class equation. But no action that I know seems to work out here: right multiplication applies when we care considering cosets; conjugation applies if one can tell the norlmalisers of the subgroups under consideration. Neither situation occurs here. Therefore I think: the proof is not included in the book, per chance it is not proved by similar means. So I come here for some help. Thanks for your attention.

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    $\begingroup$ It's not easy unless you know where to start! Are you familiar with the proof of Sylow's Theorem due to Wielandt, where you consider an action of $G$ by multiplication on the set of subsets of $G$ of order $p^b$ and use that to estimate the number of these subsets that are subgroups? This proof is usually only presented with $b=a$ but in fact it works with any $b<a$. I haven't got time to summarize it now but can do so later if no-one else helps! $\endgroup$
    – Derek Holt
    Sep 28, 2012 at 7:58
  • $\begingroup$ I know this proof, but I am not sure how to apply to this case. I must leave now: when I figure it now, I shall post it to make sure it works. Thanks for your indication. $\endgroup$
    – awllower
    Sep 28, 2012 at 8:02
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    $\begingroup$ The statement in the title is false. For example, the cyclic group of order $p^n$ has exactly $n+1$ $p$-subgroups. $\endgroup$ Sep 28, 2012 at 9:36
  • $\begingroup$ Chris, true but a single one for each order, so the statement is true! You do not look at the total number of subgroups. $\endgroup$ Sep 28, 2012 at 9:48
  • $\begingroup$ @Nicky: As I said, I was referring to the statement in the title. $\endgroup$ Sep 28, 2012 at 9:50

1 Answer 1


I agree with Derek's remark, that is a very neat way to prove it. Another way to prove it is the following and it learns you more about counting principles in $p$-groups. I assume that you have proved Exercise 1C.8 of Isaacs' book that says that you can reduce the problem to $p$-groups.

Lemma 1. Let $H$ be a proper subgroup of a finite $p$-group $G$ and suppose $|H|=p^b$. Then the number of subgroups of order $p^{b+1}$ which contain $H$, is congruent 1 mod $p$.

Proof. Fix a subgroup $H$ of order $p^b$ of the $p$-group $G$. Since $H$ is normal and proper in $N_G(H)$, one can find a subgroup $K$ with $|K:H|=p$ (look at the center of $N_G(H)/H$, which is non-trivial, pick an element of order $p$, say $\overline{x}$ and put $K=\langle x \rangle$).
On the other hand, if $K$ is a subgroup of $G$ with $H \subset K$ and $|K:H|=p$, then $H \lhd K$ (because $p$ is the smallest prime dividing $|K|$), and hence $K \subseteq N_G(H)$.
We conclude that |{$K \leq G: H \subset K$ and $|K:H|=p$}| = |{$\overline{K} \leq N_G(H)/H: |\overline{K}|=p$}|. The lemma now follows from the fact that in the group $N_G(H)/H$ the number of subgroups of order $p$ is congruent to $1$ mod $p$ (in any group, which order is divisible by the prime $p$, this is true and follows easily from the McKay proof of Cauchy’s Theorem). $\square$

Lemma 2. The number of subgroups of order $p^{b-1}$ of a $p$-group $G$ of order $p^b$ is congruent 1 mod $p$.

Proof. Let $G$ be a non-trivial group of order $p^b$. Let $\mathcal{S}$ be the set of all subgroups of $G$ of order $p^{b-1}$. Observe that this set is non-empty and fix an $H \in \mathcal{S}$. We are going to do some counting on the set $\mathcal{S}$, by defining an equivalence relation $\sim$ as follows: $K\sim L$ iff $H \cap K=H \cap L$ for $K, L \in \mathcal{S}$.
Observe that for $K, L \in \mathcal{S}$ with $K \neq L$, $G=KL$, $K \cap L \lhd G$ and $|K \cap L|=p^{b-2}$. It is easy to see that the equivalence class $[H]$ is a singleton. In addition, $H=K$ iff $H \in [K]$ for $K\in \mathcal{S}$.
Now fix a $K \in \mathcal{S}$, $K \neq H$. Counting orders one can see that if $L \in \mathcal{S}$, then $H \cap K\subset L$ iff $L \in [K]$. Hence the number of elements in $[K]$ is exactly the number of subgroups of order $p^{b-1}$ containing $H \cap K$, minus 1, namely $H$. Owing to the previous lemma, we conclude that $|[K]| \equiv 0$ mod $p$. The lemma now follows. $\square$.

Using these two lemma's you now can prove the statement by cleverly counting the number of subgroup pairs $H$ and $K$ of $G$ having order $p^b$ and $p^{b+1}$ respectively. Now back to your original question

Theorem. Let $G$ be a group of order $p^a$, $p$ prime. Let $0 \leq b \leq a$ and $n_b=$|{$H \leq G: |H|=p^b$}|. Then $n_b \equiv 1$ mod $p$.

Proof. Let $H$, $K \leq G$, with $|H|=p^b$ and $|K|=p^{b+1}$. Define a function $f$ as follows: $f(H,K)=1$ if $H \subset K$ and $f(H,K)=0$ otherwise. Let us compute $\sum_{H} \sum_{K} f(H,K)$ in two different ways: $\sum_{H}$$\sum_{K} f(H,K) = \sum_{H} \sum_{H \subset K}1$ $\equiv \sum_{H} 1$ mod $p$, according to Lemma 1 above. Similarly, by reversing the order of summation of $H$ and $K$, the sum equals $\sum_{K} 1$ mod $p$, using Lemma 2. In other words, for all $b$, the number of subgroups of $G$ of order $p^b$ is congruent mod $p$ to the number of subgroups of $G$ of order $p^{b+1}$. The theorem now follows from the fact that the number of subgroups of order $p^a$ equals 1, namely $G$ itself. $\square$.

Remark. The theorem counts all subgroups of fixed order $p^b$. If we restrict ourselves to the normal subgroups of order $p^b$ the same holds: |{$H \unlhd G: |H|=p^b$}|$\equiv 1$ mod $p$. Sketch of proof: let $G$ act by conjugation on the set of all subgroups $H$ of order ${p^b}$. The fixed point are exactly the normal subgroups. Now apply the Theorem above.

Finally, also see this StackExchange entry.

  • $\begingroup$ Thanks!Just when I am not here, I found exactly the same proof!It is indeed interesting, and thanks for pointing out here. However, I used the exercise of the same book to show that the number of subgroups of order p is congruent to 1, and then use an induction. $\endgroup$
    – awllower
    Sep 28, 2012 at 10:09
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    $\begingroup$ A nice way to think about this "clever" counting is to arrange a matrix with rows corresponding to subgroups of order $p^b$ and columns corresponding to subgroups of order $p^{b+1}$. Then put a $1$ in $(i,j)$ if subgroup $j$ contains subgroup $i$, and a $0$ otherwise. Now use row sums and column sums... $\endgroup$
    – user641
    Sep 28, 2012 at 11:46
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    $\begingroup$ Steve, yes that is a very elegant way of proving it. Like the way of proving the orbit counting formula of Burnside/Cauchy/Frobenius! $\endgroup$ Sep 28, 2012 at 14:42
  • $\begingroup$ May I ask for the details? I am waiting for it to be accepted. In any case, if it is too troublesome, I can just accept it.Regards. $\endgroup$
    – awllower
    Feb 15, 2013 at 5:43
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    $\begingroup$ @awllower Thanks. Yes in combinatorial/discrete mathematics this phenomenon of summing in different ways often appears. It maybe warrants a new entry in this StackExchange to ask for examples ... $\endgroup$ Mar 8, 2013 at 13:27

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