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Since yesterday, I've tried to give an answer to this question.

Let $n$ be a positive natural number and $p$ a prime. The group $(\mathbb{Z}/p^n\mathbb{Z})^\times$ will be denoted by $G_n$.

Let $f : G_n \rightarrow G_n,\ x \mapsto x^{p-1}$ and $g : G_n \rightarrow G_n,\ x\mapsto x^{p^{n-1}}$. Then $f\circ g$ gives the trivial endomorphism of $G_n$ since we reach $\phi(p^n)$, which means $g[G_n] \subseteq \ker (f)$. But actually are they equal or not ?

I wonder if the Snake Lemma could be helpful in some way...

This is a real question I'd love to know the answer of. Thank you!

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    Well, if $p$ is odd, then $G_n$ is cyclic, say with a generator $a$, so the kernel of $f$ must consists of $a^{p^{n-1}t}$ for some $t=0,1,\ldots,p-2$. – 2017-02-22
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    If $p$ is odd then $(\mathbb Z/p^n\mathbb Z)^\times \cong \mathbb Z/(p-1)\mathbb Z\times \mathbb Z/p^{n-1}\mathbb Z$ by the Chinese Remainder Theorem. $f$ acts as zero on the first and as an isomorphism on the second factor, whereas $g$ acts as an isomorphism on the first and as zero on the second factor. Hence the image of $g$ coincides with the kernel of $f$ (and vice versa). – 2017-02-22
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    Is the purpose of this exercise to prove that $G_n$ is cyclic when $p$ is odd? – 2017-02-22

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