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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

the_fox
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HeMan
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  • It is only a surjective homomorphism. – HeMan Jul 15 '19 at 05:02
  • Something is a bit off. By the first isomorphism theorem $\mathrm{Ker}(\sigma)$ has order $p$ (not index!!), so when $m\ge3$ there are certainly non-trivial subgroups $H$ not contained in the kernel. – Jyrki Lahtonen Jul 15 '19 at 05:06
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    It's not true. Consider the case $m=3$, and the subgroup $H$ of order $p^2$. – Angina Seng Jul 15 '19 at 05:06
  • May be the codomain should be $C_p$? – Jyrki Lahtonen Jul 15 '19 at 05:08
  • Right, sorry. I got carried away. – the_fox Jul 15 '19 at 05:09
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    To get the same conclusion, you need the target group to have order $p$. Not sure what else the author could have meant here. The only other thing I can imagine is (with domain and target as stated) that every non-trivial subgroup contains the kernel. – the_fox Jul 15 '19 at 05:09
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    I have had a look at the relevant section of Robert's book. He is mistaken. The $C_{p^{m-1}}$ is not a typo because it is used explicitly more than once. (At the end of his proof, he writes explicitly that the kernel has order $p^{m-1}$.) – the_fox Jul 15 '19 at 05:21

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