Usbrth

Is it true that $Z_p^2^* simeq Z_ptimes Z_p-1$? One can verify that $|Z_p^2^*|=p(p-1)$. Can you give an isomorphism?

Any two cyclic groups of the same order are isomorphic. $mathbbZ_ptimesmathbbZ_p-1$ is cyclic by the Chinese remainder theorem while $mathbbZ_p^2^*$ is cyclic because there is a primitive root mod a power of a prime number. (this is a theorem in number theory)

Here is a sketch for an odd prime:

• 
  $1+p$ has order $pbmod p^2$.


• 
  $mathbf Z_p^times $ is cyclic, of order $p-1$, hence, if $a bmod p$ is one of its generators, $abmod p^2$ has order a multiple of $p-1$. So a power $b$ of $a$ has order exactly $p-1bmod p^2$.


• Thus , as $p$ and $p-1$ are coprime, $(1+p)b$ has order $p(p-1)$.

Now, let $u$ be a unit $bmod p^2$. It can be written as $u=(1+p)^r b^s$ $;(0le r<p,: 0le s <p-1)$.
This enables us to define an isomorphism as follows:
beginalign
mathbf Z_p^2^times & longrightarrow mathbf Z_ptimesmathbf Z_p-1 \
u=(1+p)^r b^s&longmapsto(rbmod p, sbmod p-1)
endalign

The operation of $mathbbZ^*_p^2$ is multiplication and the operation of $mathbbZ/pmathbbZ times mathbbZ/(p-1)mathbbZ $ of component addition. It remains to find a map $phi: mathbbZ^*_p^2 rightarrow mathbbZ/pmathbbZ times mathbbZ/(p-1)mathbbZ$ that satisfies $phi(ab) = phi(a) + phi (b)$ or show that no map exists.

Case example: $ p = 2 $

Let us first take $a=1$. Then, $phi(b) = phi(1)+ phi(b) implies phi(1) = 0$ (here, $0$ is $(0,0)$). Thus, $phi(3) = (1,0)$. This is easily verified to be a homomorphism under the group operations. But the inverse map is not a homomorphism: $phi^-1((1,0)+(0,0)) = 3 neq phi^-1(0,0) + phi^-1(1,0)$.

Your claim is not true for the even primes. For the odd primes, we may discuss the work of user Bernard.

The canonical surjective ring homomorphism $mathbf Z/p^2 to mathbf Z/p$ defined by $a$ mod $p^2 to a$ mod $p$ obviously induces a group homomorphism $ f :(mathbf Z/p^2)^* to (mathbf Z/p)^*$ which is surjective because any integer $a$ representing a class of $mathbf Z/p$ and prime to $p$ will represent a class of $(mathbf Z/p^2)^*$. So Imf$f=(mathbf Z/p)^*$ is cyclic of order $(p-1)$ (every finite multiplicative subgroup of a field is cyclic), hence is isomorphic to $(mathbf Z/(p-1), +)$.

Let us determine Ker$f$. If an integer $a$ is $equiv 1$ mod $p$, it can be written $equiv 1+kp$ mod $p^2mathbf Z$. Letting $k=0,1,...,p-1$, one gets $p$ distinct classes of $(mathbf Z/p^2)^*$ which are all in Ker$f$. This shows at the same time that Ker$f$ is the image of the multiplicative group $1+pmathbf Z$ in $mathbf Z/p^2mathbf Z$ and has order $p$, hence is isomorphic to $(mathbf Z/p, +)$. Since $p$ and $(p-1)$ are coprime, the CRT actually gives a direct product decomposition $((mathbf Z/p^2)^*,times)cong (mathbf Z/p, +)times (mathbf Z/(p-1), +)$.

NB : This argument can be extended almost without change to show that, for any odd prime $p$ and any $rge 1$, $(mathbf Z/p^r)^*$ is cyclic, $cong (mathbf Z/p^r-1)times (mathbf Z/(p-1))$. The prime $2$ plays the usual trouble maker : for $rge 3, (mathbf Z/2^r)^*$ is abelian of type $(2^r-2, 2)$. Note that these results are sharper than the usual Euler totient function which gives the order of $(mathbf Z/p^r)^*$.