Another proof of Strickland's result.

Let $G$ be a group of order $p^n$ and $\nu(G)$ be the sum of orders of its cyclic subgroups. To prove that $\nu(G)=\sigma_1(G)$, we proceed by induction on $|G|$. If $G$ is cyclic, then, obviously, $\nu(G)=\sigma_1(G)$. Therefore assume that $G$ is noncyclic. If $H<G$, then, by induction, $\nu(H)=\sigma_1(H)$ depends only on $|H|$. Therefore assume that $G$ is noncyclic. Then it has a normal subgroup $T$ such that $G/T$ is elementary abelian of order $p^2$. In that case, as all maximal subgroups of $G$ have the same order, we get, provided $H>T$ is a subgroup of index $p$ in $G$, then
$$
\nu(G)=(p+1)\nu(H)-p\nu(T)
$$
hence, by induction,
$$
\nu(G)=(p+1)\sigma_1(p^{n-1})-p\sigma_1(p^{n-2})
$$
$$
=(p+1)(1+p+\dots+p^{n-1})-p(1+p+\dots+p^{n-2})=1+p+\dots+p^n=\sigma_1(p^n),
$$
completing the proof.

Above we applied a partial case of Hall's enumeration principle. Applying that principle, it is easy to compute, for not very complicated $p$-groups, the sum of orders of their subgroups. About Hall's enumeration principle, see Kap. III in Huppert's `Endliche Gruppen'.