$H$ is a cyclic group of order $p^m$.

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I am reading "A Course in $p$-adic Analysis" by Alain M. Robert. I have found the following statement (page 60):

Let $\sigma \colon C_{p^m} \rightarrow C_{p^{m-1}}$ be a surjective homomorphism between cyclic groups of order $p^m$ and $p^{m-1}$, respectively. If $H$ is a subgroup of $C_{p^m}$ not contained in $\ker(\sigma)$, then $H=C_{p^m}$.

It basically says that as $\phi(p^m)= p^m - p^{m-1}$ and the order of the kernel is $p^{m-1}$, where $\phi$ is Euler's function, the result follows, but I really don't understand why. I think that by Langrange and the first isomorphism theorem, the order of the kernel is just $p$. So I'm very confused. Do you have any hint or solution?

Thanks in advance.enter image description here