Let $G$ be a p-group with $|G|=p^n$. Show $G$ has normal subgroups of order $p^m$ for each $0Here is what I have so far. 
Statement: Let $G$ be a p-group with $|G|=p^n$. Then $G$ has normal subgroups of order $p^m$ for each $0<m<n$.
The case $|G|=p^1$ is trivial. Proceed by induction on $n$ (where $|G|=p^n$), and assume the statement holds for all groups of order $|G|=p^m$ such that $m<n$. 
It follows from a theorem due to Burnside (the center of any p-group is nontrivial) that there exists an element $a\in Z(G)$ of order $p$. Clearly then $|G|/\left<a\right>$ is a group of order $p^{n-1}$, and by the induction hypothesis, $|G|/\left<a\right>$ contains normal subgroups of orders $\,p,\, p^2,\dots,\,p^{n-2}$ (note that it also contains a normal subgroup of order $p^{n-1}$, trivially).
This is where I am stuck. I know that there is a 1-1 correspondence 
$$\{\text{subgroups of }|G|/\left<a\right>\}\longleftrightarrow\{\text{subgroups }H \text{ of } G\text{ which contain }\left<a\right>\},$$ and that normal subgroups correspond to normal subgroups under this correspondence. What I can't figure out is how to say anything about the order of these subgroups $H < G$ which contain $\left<a\right>$. I understand that each $H$ has order a multiple of $p$, but I can't nail down why exactly there must exist one of each of the orders $\,p,\, p^2,\dots,\,p^{n-1}$. Couldn't $G$ contain two distinct subgroups of order $p^2$ which contain $\left<a\right>$, for example?
Any help would be sincerely appreciated, thank you.

That 1-1 correspondende preserve indexes, that is $(G:H) = (G/<a>:H')$, where $H'$ is the subgroup of $G/<a>$ corresponding to $H$. This group $H'$ can be identified with $H/<a>$ and therefore preservation of indexes follows from Lagrange's theorem for instance.
