 Solution

Put $q=p^n$ and choose a generator $h$ of $H$. In characteristic $p$, $(h-1{)}^q=h^q-1=0$, soA module is a vector space with a nilpotent operator $t$ whose nilpotence index is at most $q$. The nilpotent Jordan normal form, valid over every field, splits it into Jordan blocks. One block of size $i$ isIt is indecomposable: its endomorphism ring is $k[t]/(t^i)$, a local ring with no nontrivial idempotents. Distinct $i$ have distinct dimensions. Conversely a direct sum of at least two Jordan blocks is decomposable. Thus there are exactly $q$ indecomposable isomorphism classes.

A homomorphism $V_i\to V_j$ is determined by the image of $1$, which can be any element of $V_j$ annihilated by $t^i$. If $i\ge j$ this is all of $V_j$, of dimension $j$; if $i<j$ it is the span of $t^{j-i},\dots ,t^{j-1}$, of dimension $i$. HenceThis realizes the indecomposable modules for a cyclic p-group.