In $p$-groups subgroup of order $1$ less are always normal Prove that any subgroup of order $p^{n-1}$ in a group $G$ of order $p^n$, $p$ a prime number, is normal in $G$.
Remark: I have met the similar problem here but I would like to complete my solution which is different.
Proof: $G$ be a group of order $p^n$ and $H$ it's  subgroup of order $p^{n-1}
$. Let $S=\{Hx: x\in G\}$ the set of right cosets of $H$ in $G$. Then $o(S)=i_G(H)=p$. 
Then it's easy to show that exists $\psi:G\to S_p$ where $\psi$ is homomorphism and $K=\text{Ker} \psi$ is the largest normal subgroup of $G$, contained in $H$.
Let $f=\psi \mid_H$ is the mapping $\psi$ restricted to $H$ then consider $f:H\to S_p$ then $f(H)$ is a subgroup of $S_p$. Thus, $o(f(H))\mid p!$
Also $H/\text{Ker} f \cong f(H) $ but $\text{Ker} f=\text{Ker} \psi \cap H=\text{Ker} \psi=K$ $\Rightarrow$ $\dfrac{o(H)}{o(K)}=o(f(H))$ $\Rightarrow$ $\dfrac{p^{n-1}}{o(K)}=o(f(H))\mid p!$
So we have two cases $o(f(H))=p^{\alpha}$ where $\alpha\in \{0,1\}$
1) If $\alpha=0$ then $o(f(H))=1$ so $o(K)=o(H)$ and $K\subset H$ $\Rightarrow$ $K=H$ where $H$ is normal in $G$ since kernel is always normal subgroup.
2) If $\alpha=1$ then $o(f(H))=p$ and I don't know how to complete this case.
Would be very grateful if anybody can show how to do in the second case.
 A: As you say there is a homomorphism $\psi$ from $G$ to $S_p$ defined by its
action on cosets of $H$. The image of such a map is transitive, and
the order of the image is a power of $p$. Thus $|\psi(H)|=p$. The image of $\psi$ has order $p$, and is a cyclic group.
The kernel of $\psi$ must be $H$, and so $H$ is normal in $G$.
A: Let me clarify a few things. 
The map $\Psi$ is defined by $\Psi(g):aH\mapsto gaH$. An element in Ker$\Psi$ has to fix every single coset, in particular $H$ itself: $$g\in \operatorname{Ker}\Psi\Longrightarrow gH=H\Longrightarrow g\in H$$
Therefore $\operatorname{Ker}\Psi\subset H$.
A consequence of this is that $\operatorname{Ker}\Psi$ is not the whole group $G$.
Now look at $G/\operatorname{Ker}\Psi\simeq \operatorname{Im}\Psi$. Since $G$ is a $p$-group, and since $\operatorname{Ker}\Psi\neq G$, we know that $$|\operatorname{Im}\Psi|=p^a$$
for some positive $a$. But we also know that $p^a$ has to divide the order of $\mathfrak{S}_p$, therefore $a=1$.
It follows that $$\dfrac{|G|}{|\!\operatorname{Ker}\Psi|}=p$$
Therefore $$|\!\operatorname{Ker}\Psi|=p^{n-1}=|H|$$
And since $\operatorname{Ker}\Psi\subset H$, we have equality $\operatorname{Ker}\Psi= H$.
Finally, since $H$ is a kernel, $H$ is normal.
