$p^k\mid |G|$ for $k=1,\dots,\alpha$, and $n_k=\#$ of subgroups of order $p^k$. Is $n_k\equiv 1\pmod p$ for every $k=1,\dots,\alpha$? Let $p$ be a prime divisor of the order of a finite group $G$, and $\alpha$ the greatest power of $p$ dividing $|G|$. Let $n_k$ be the number of subgroups of $G$ of order $p^k$ (so, $n_\alpha$ is the number of $p$-Sylow subgroups of $G$, usually referred to as "$n_p$"). The following three facts are compatible with the subsequent (conjectured) claim:

*

*$n_\alpha\equiv 1\pmod p$, by Sylow III;

*$n_k\ge 1$, for every $k=1,\dots,\alpha$: this is, e.g., the Theorem 2.12.1 in Herstein's Topics in algebra;

*$n_1\equiv 1\pmod p$: this is a corollary of this result: in fact, in the notation of the link, $p\mid |X|=n_1(p-1)+1=n_1p-(n_1-1)$ if and only if $p\mid n_1-1$ if and only if $n_1\equiv 1\pmod p$.

Claim. $n_k\equiv 1\pmod p$ for every $k=1,\dots,\alpha$.
According to this and this, the claim holds true for the particular cases $|G|=2^33$ and $|G|=2^43$, respectively.
Q: Is the claim true?
 A: Let us prove the following (this is Problem 1C.8 in M.I. Isaacs Finite Group Theory).
Theorem Let $P \in Syl_p(G)$ and $a$ a non-negative integer. Let
$$\Gamma=\{ \text{subgroups of $G$ of order } p^a\}$$
$$\Pi=\{ \text{subgroups of $P$ of order } p^a\}.$$
Then $\#\Gamma \equiv \#\Pi$ mod $p$.
Now, before proving this theorem, let's see how it follows that if $p^a$ divides $|G|$ ($p^a$ not necessarily being the largest $p$-power dividing $|G|$), the number of subgroups of order $p^a$ of $G$ is $\equiv 1$ mod $p$. 
If ones takes $a$ in the theorem above such that $p^a$ is the maximum power, clearly $\#\Pi=1$ and $\#\Gamma = n_p(G)$, the number of Sylow $p$-subgroups : one recovers the third Sylow theorem. Further, in general, if $P$ is a $p$-group and $p^a \mid |P|$, then the number of subgroups of order $p^a$ is $\equiv 1$ mod $p$, see for proofs here. These two facts combined with the theorem above gives the required result: if $G$ is a finite group and $p^a \mid |G|$, then the number of $p$-subgroups of order $p^a$ is $\equiv 1$ mod $p$.
Now we need a small lemma.
Lemma Let $P \in Syl_p(G)$ and $Q$ a $p$-subgroup of $G$. Then the following are equivalent.
(a) $Q \unlhd P$
(b) $P \subseteq N_G(Q)$.
Proof (a) $\Rightarrow$ (b) is obvious: since $Q$ is normal in $P$, $P$ normalizes $Q$.
(b) $\Rightarrow$ (a). Since $P \in Syl_p(G)$ and $P \subseteq N_G(Q)$, certainly $P \in Syl_p(N_G(Q))$. But of course, $Q \unlhd N_G(Q)$ and, since Sylow subgroups are conjugate, $Q$ is contained in every Sylow $p$-subgroup of $N_G(Q)$. $\square$
Proof of the Theorem: $P$ acts on $\Gamma$ by conjugation, let $\Omega$ be the set of fixed-points. Then $Q \in \Omega$ if and only if $P \subseteq N_G(Q)$, which by the Lemma is equivalent to $Q \unlhd P$. Hence $\Omega=\{Q \unlhd P: |Q|=p^a\}$. By the Orbit-Stabilizer Theorem and the fact that $P$ is a $p$-group, we conclude $\#\Gamma \equiv \#\Omega$ mod $p$.
$P$ also acts on $\Pi$ by conjugation and the fixed points for this action are exactly the elements of the set $\Omega$. So, $\#\Pi \equiv \#\Omega$ mod $p$, yielding $\#\Gamma \equiv \#\Pi$ mod $p$. $\square$ 
Final observation We now know that if $G$ is a finite group and $p^a \mid |G|$, then the number of $p$-subgroups of order $p^a$ is $\equiv 1$ mod $p$. From this it equally follows that if $G$ is a finite group and $p^a \mid |G|$, then the number of normal $p$-subgroups of order $p^a$ is $\equiv 1$ mod $p$. Can you prove this?
A: This is a theorem of Frobenius proved in [F. G. Frobenius, Verallgemeinerung des Sylow’schen Satzes, Sitzungsber. Preuß. Akad. Wiss. 1895 (1895), 981–993].
The most elegant proof is probably the one by Wielandt, see [Huppert, Satz I.7.2] or https://math.stackexchange.com/a/479869/960602.
