# Number of normal subgroups of order $p^s$ of a $p$-group

Let $$G$$ be a $$p$$-group. Show that the number of normal subgroups of $$G$$ that have order $$p^s$$ is $$1$$ $$mod(p)$$.

I think I have to use Sylow theorems and the fact that every subgroup of order $$p^s$$ is contained in a $$p$$-Sylow subgroup, but I don't know how to apply this. I also thought about defining an action in the set of normal subgroups of $$G$$ of order $$p^s$$, but I couldn't think of one that is useful.

Any help will be very appreciated. Thanks!

This is not so trivial. You first show the statement for all subgroups of a particular order, so not necessarily normal subgroups. This follows from subtle counting and see the final remark for answering your question, using the Orbit-Stabilizer Theorem.

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 points are exactly the normal subgroups. Now apply the Theorem above.

As an alternative to the proof given by Nicky Hekster, I would suggest using Wielandt's proof of Sylow's theorem to prove that the total number of subgroups of $$G$$ of order $$p^s$$ is $$1 \bmod p$$.

Then consider the conjugation action of $$G$$ on all such subgroups. The non-normal subgroups are in orbits with lengths divisible by $$p$$, so the number of normal subgroups is also $$1 \bmod p$$.