# The multiplicative group of units in $\mathbb{Z}_{p}$ is isomorphic to $\mathbb{Z}_{p} \times C_{n}$ with $n=\max\{p-1,2\}$.

Problem. (a) Write down explicitly the rules for addition and multiplication in $$\mathbb{Z}_{p}$$.

(b) Show that $$(\mathbb{Z}_{p^{i}},\varphi_{ij})$$ where $$\varphi:\mathbb{Z}_{p^{j}} \to \mathbb{Z}_{p^{i}}$$ is given by $$\varphi_{ij}(n + p^{j}\mathbb{Z}) = n + p^{i}\mathbb{Z}$$ is a inverse system of finite rings and groups.

(c) Show that $$\mathbb{Z}_{p} = \varprojlim_{i}\mathbb{Z}_{p^{i}}.$$

(d) Show that the multiplicative group of units in $$\mathbb{Z}_{p}$$ is isomorphic to the direct product of $$\mathbb{Z}_{p}$$ with a cyclic group of order $$\max\{p-1,2\}$$.

I have no problems with (a), (b) and (c).

For item (d), I dont know how to approach this problem. I take an arbitrary cylic group of order $$\max\{p-1,2\}$$ and I tried to construct a isomorphism between the groups, but it doesn't works (at least, I cannot see how to do that). I would like some hints and approaches to follow.

Also, in the item (c), I used the inverse system given in (b) and I built a bijection between $$\mathbb{Z}_{p}$$ and $$\varprojlim \mathbb{Z}_{p^{i}}$$, considering $$\varprojlim \mathbb{Z}_{p^{i}}$$ as a subgroup of $$\prod_{i}\mathbb{Z}_{p^{i}}$$. Generally, this is the standard approach to problems like that. But I would like to know any other alternative approach.

Edit. I read the references in comment below, but it uses some things that the book doesn't comment on (like exact sequences, Hensel's lemma, for example). Basically, the book defines $$\mathbb{Z}_{p}$$ as the set of formal infinite sums and proves only the necessary results to find the completion of $$\mathbb{Z}$$, I mean: the book doesn't develop until now $$p$$-adic theory. So, I would like to know if there is another proof.

• I think this is answered here together with here, (although there may be a bit more work to do when $p=2$). – Derek Holt Apr 26 at 12:02
• Thank you, @DerekHolt! I will check the links! – Lucas Corrêa Apr 29 at 23:07
• With the restrictions imposed on what methods are allowed, the natural thing to do is to compute the unit groups of the finite rings $\Bbb Z/p^j \Bbb Z$ which make up the inverse limit (and for which, by the way, I find "$\Bbb Z_{p^j}$" a horrible notation -- what book is that from?). – Torsten Schoeneberg May 7 at 18:12
• @TorstenSchoeneberg let me see if I got your idea. I should to show that the unit group of $\mathbb{Z}/p^j \mathbb{Z}$ is isomorphic to something and when I calculate the inverse limit of something, the answer will be $\mathbb{Z}_{p} \times C_{n}$? – Lucas Corrêa May 9 at 14:37
• @TorstenSchoeneberg, this question was rewritten for a friend, but its from Wilson's book "Profinite Groups", but Wilson uses $\mathbb{Z}/p^j \mathbb{Z}$. – Lucas Corrêa May 9 at 14:40

I'll follow up my comment with a sketch of how I assume the exercise is meant to be solved. People with knowledge in $$p$$-adics will see I try to recover the standard (for $$p \neq 2$$) $$\Bbb Z_p^\times \simeq \mu(\Bbb Q_p) \times (1+p\Bbb Z_p)$$ -- and Teichmüller representatives/roots of unity on the first factor, and the second factor being $$\simeq (1+p)^{\Bbb Z_p} \simeq (\Bbb Z_p, +)$$ -- with arithmetic congruences modulo $$p$$-powers.

Let's look at the rings $$\Bbb Z/p^j\Bbb Z$$. Everyone since Euler knows that the unit group $$(\Bbb Z/p^j\Bbb Z)^\times$$ has order $$(p-1)\cdot p^{j-1}$$, which suggests it's of the form $$C_{p-1} \times C_{p^{j-1}}$$, where $$C_n$$ denotes the cyclic group of order $$n$$. (That's actually not the case for $$p=2$$ ($$j\ge 3$$), see below, but indeed for all other primes.) Well great: If that is the case, and furthermore if we can establish these isomorphisms in a compatible way when we let $$j$$ vary, then we have $$\Bbb Z_p^\times \simeq\varprojlim (\Bbb Z/p^j\Bbb Z)^\times \simeq \varprojlim (C_{p-1} \times C_{p^j}) \simeq \varprojlim C_{p-1} \times \varprojlim C_{p^j} \simeq C_{p-1} \times \Bbb Z_p$$ (there's a few things to check here maybe, but this chain of isomorphisms should be straightforward).

For $$p \neq 2$$, such compatible isomorphisms $$(\Bbb Z/p^j\Bbb Z)^\times \simeq C_{p-1} \times C_{p^{j-1}}$$can be found like this:

• First, the $$C_{p^{j-1}}$$-part. I claim that the element $$1+p$$ (formally, its various residues mod $$p^j$$) generates such a cyclic subgroup. Namely, Let $$y_j=$$ the residue of $$1+p$$ in $$\Bbb Z/p^j\Bbb Z$$.

Check that the order of $$y_j$$ in $$(\Bbb Z/p^j\Bbb Z)^\times$$ is $$p^{j-1}$$, i.e. that $$(1+p)^{p^k} \not \equiv 1$$ mod $$p^{j}$$ for $$k , but $$(1+p)^{p^{j-1}} \equiv 1$$ mod $$p^{j}$$. For this, you need some playing with $$p$$-divisibility of binomial coefficients $$\binom{p^k}{n}$$, and at some point you really need $$p \neq 2$$.

Compatibility is clear, as the projection $$(\Bbb Z/p^j\Bbb Z) \rightarrow (\Bbb Z/p^i\Bbb Z)$$ sends $$y_j$$ to $$y_i$$.

• The $$C_{p-1}$$ part needs different considerations. A crucial fact here, proved by induction, is $$a \equiv b \text{ mod } p \Rightarrow a^{p^{j-1}} \equiv b^{p^{j-1}} \text{ mod } p^j \quad (*)$$ With this, for $$a \in\lbrace 1, ..., p-1\rbrace$$, check that the residues of $$a^{p^{j-1}}$$ in $$\Bbb Z/p^j\Bbb Z$$ are distinct from each other and form a subgroup isomorphic to $$(\Bbb Z/p\Bbb Z)^\times$$, which is cyclic of order $$p-1$$.

Choose and fix $$2 \le a \le p-1$$ such that its residue mod $$p$$ is a generator of $$(\Bbb Z/p\Bbb Z)^\times$$. Then $$x_j :=$$ the residue of $$a^{p^{j-1}}$$ in $$(\Bbb Z/p^j\Bbb Z)^\times$$ is a generator of the $$C_{p-1}$$ we want, and by the fact $$(*)$$ and iterations of Fermat's little theorem, we also have that the projection $$(\Bbb Z/p^j\Bbb Z) \rightarrow (\Bbb Z/p^i\Bbb Z)$$ sends $$x_j$$ to $$x_i$$.

Finally, for $$p=2$$, the unit group turns out to be not $$C_{2^{j-1}}$$, but $$C_2 \times C_{2^{j-2}}$$ (for $$j \ge 3$$). To adapt the above, for $$y_j$$ instead of $$1+p$$ take $$1+p^2$$ (i.e. $$5$$) mod $$2^j$$, and show its order in the unit group is $$2^{j-2}$$. Funnily, here the constant $$C_2$$ part is easy, namely it's generated by $$x_j :=$$ the residue of $$-1$$ in $$(\Bbb Z/2^j\Bbb Z)$$.

• Thank you for this great answer! I will check the details with attention, but it seems very clear to me! – Lucas Corrêa May 13 at 19:57