What's an isomorphism between $Z_{p^2}^*$ and $Z_p\times Z_{p-1}$

Question

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

Answer 1

Any two cyclic groups of the same order are isomorphic. $\mathbb{Z_p}\times\mathbb{Z_{p-1}}$ is cyclic by the Chinese remainder theorem while $\mathbb{Z_{p^2}^*}$ is cyclic because there is a primitive root mod a power of a prime number. (this is a theorem in number theory)

Answer 2

Here is a sketch for an odd prime:

• $1+p$ has order $p\bmod p^2$.
• $\mathbf Z_p^\times $ is cyclic, of order $p-1$, hence, if $a \bmod p$ is one of its generators, $a\bmod p^2$ has order a multiple of $p-1$. So a power $b$ of $a$ has order exactly $p-1\bmod p^2$.
• 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 $u=(1+p)^r b^s$ $\;(0\le r<p,\: 0\le s <p-1)$. This enables us to define an isomorphism as follows: \begin{align} \mathbf Z_{p^2}^\times & \longrightarrow \mathbf Z_p\times\mathbf Z_{p-1} \\ u=(1+p)^r b^s&\longmapsto(r\bmod p, s\bmod p-1) \end{align}

Answer 3

The operation of $\mathbb{Z}^*_{p^2}$ is multiplication and the operation of $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/(p-1)\mathbb{Z} $ of component addition. It remains to find a map $\phi: \mathbb{Z}^*_{p^2} \rightarrow \mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/(p-1)\mathbb{Z}$ 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.

Answer 4

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 $p^2\mathbf Z$. 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 $1+p\mathbf Z$ in $\mathbf Z/p^2\mathbf Z$ 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 $r\ge 1$, $(\mathbf Z/p^r)^*$ is cyclic, $\cong (\mathbf Z/p^{r-1})\times (\mathbf Z/(p-1))$. The prime $2$ plays the usual trouble maker : for $r\ge 3, (\mathbf Z/2^r)^*$ 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 $(\mathbf Z/p^r)^*$.