  We prove that a Volterra-type integral operator $T_gf(z) = \int_0^z
f(\zeta)g'(\zeta)d\zeta, \, z \in \mathbb D,$ defined on Hardy spaces $H^p, \,
1 \le p < \infty,$ fixes an isomorphic copy of $\ell^p,$ if the operator $T_g$
is not compact. In particular, this shows that the strict singularity of the
operator $T_g$ coincides with the compactness of the operator $T_g$ on spaces
$H^p.$ As a consequence, we obtain a new proof for the equivalence of the
compactness and the weak compactness of the operator $T_g$ on $H^1$.
