I have the following problem:$$$$ Let $p$ be a prime number and $G$ be a set of all $x\in \mathbb{Z}_{p^2}$, such that $x \equiv 1 \pmod{p}$. Prove that:
- $G$ is a multiplicative group (regarding multiplication in $\mathbb{Z}_{p^2}$).
- $|G| = p$.
- $L : G \longrightarrow \mathbb{Z}_p$, where $L(x) = \frac{x - 1}p \pmod{p}$ is a group isomorphism.
- $G$ is generated by $p+1$ and $L$ is a discrete logarithm with base $p+1$ in G. In other words prove that $(p+1)^{L(x)}\pmod{p^2} = x$ for every $x$.
I can prove the first three points but I'm having trouble with the last one. I've found that the Fermat–Euler theorem is applicable, so we get that $$(p+1)^{\phi(p^2)}\equiv1\pmod{p^2}$$ $$(p+1)^{p(p-1)}\equiv1\pmod{p^2}$$ $$x(p+1)^{p(p-1)}\equiv x\pmod{p^2}$$ Now since $x \equiv 1\pmod{p}$ we can deduce that $x = 1 + pk$ for some $k$. Substituting in the above expression we get $(1 + pk)(p + 1)^{p(p-1)}\equiv (1 + pk)$ . On the other hand we want $$(p+1)^{\frac{x-1}p} \equiv x \pmod{p^2}$$ $$(p+1)^{\frac{1 + pk - 1}p} \equiv (1 + pk) \pmod{p^2}$$ $$(p+1)^k \equiv (1 + pk) \pmod{p^2}$$ So we just need to prove that $(1 + pk)(p + 1)^{p(p-1)} \equiv (p+1)^k \pmod{p^2}$ which I cant't do. I'm pretty sure that there is a much easier way to prove that $L$ is a discrete logarithm but I have no idea how. Any help would be greatly appreciated.
We know that $x\equiv 1+kp\pmod{p^2}$ for some integer $k$ that is specified modulo $p$. Furthermore, we decided to denote $k=L(x)$.
The last item amounts to showing that $$ (1+p)^k\equiv 1+kp \pmod{p^2}. $$ This follows from the binomial formula $$ (1+p)^k=\sum_{i=0}^k\binom k i p^i $$ by observing that the terms with $i>1$ are all divisible modulo $p^2$ (and the known fact that $\binom k 1 =k$).