In the group $G$ of units of $\mathbb Z_{mp^k}$, the subset $\{g \in G | \operatorname{int}(g) \operatorname{mod}m = 1\}$ forms a cyclic subgroup

I want to prove, using only elementary theorems, that in the group $$G$$ of units of $$\mathbb Z_{mp^k}$$, where $$p$$ is an odd prime, $$k$$ a positive integer and $$m$$ a positive integer such that $$(p^k,m) = 1$$, the subset $$H =\{g \in G | \operatorname{int}(g) \operatorname{mod}m = 1\}$$ forms a cyclic subgroup of order $$(p-1)p^{k-1}$$.

Let $$g \in G$$ then $$g$$ can be written as $$am + b$$ where $$b = \operatorname{int}(g) \operatorname{mod} m$$ and $$a = (\operatorname{int}(g)-b)/m$$. In this way an element $$g \in G$$ can be assigned "coordinates" $$[a,b]$$. The subset $$H$$ in consideration can then be seen as the subset with coordinates $$[a,1]$$. Example: with $$p = 5, k = 2$$ and $$m = 4$$ the coordinates of this subset are: $$[ 0, 1 ], [ 2, 1 ], [ 3, 1 ], [ 4, 1 ], [ 5, 1 ], [ 7, 1 ], [ 8, 1 ], [ 9, 1 ], [ 10, 1 ], [ 12, 1 ], [ 13, 1 ], [ 14, 1 ], [ 15, 1 ], [ 17, 1 ], [ 18, 1 ], [ 19, 1 ], [ 20, 1 ], [ 22, 1 ], [ 23, 1 ], [ 24, 1 ]$$

It is not hard to prove that $$H$$ forms a subgroup of $$G$$. The coordinates "$$a$$" occuring in $$H$$ all satisfy $$\gcd(am+1, mp^k)=1$$. Since $$am+1$$ and $$m$$ can have no common non trivial factor this condition is equivalent to $$\gcd(am+1, p^k)=1$$. From Bézouts lemma one can find $$u,v$$ such that $$up^k+vm=1$$, substituting $$1$$ then gives the condition $$\gcd(am+up^k+vm,p^k)=1 \Leftrightarrow \gcd(am+vm,p^k)=1$$. Since $$(m,p^k)=1$$ we must have $$a \neq -v \operatorname{mod} p$$. In our example we have $$v = -6$$ so the coefficients all satisfy $$a \neq 1 \operatorname{mod} 5$$. From this we conclude that the number of $$a$$'s is $$p^k-p$$. The only thing that remains to prove is that this group is cyclic. The only thing I tried was to see how the usual multiplication on $$H$$ translate in that of the composition rule it induces on the coordinates $$a$$ giving $$a \cdot a' = maa'+a+a'\operatorname{mod} mp^k$$, but here is where I'm stuck.

• @reuns Sorry I mixed up in the title, corrected in the mean while – Marc Bogaerts Jun 12 at 16:16

Take $$b,c$$ such that $$bp^k \equiv 1 \bmod m$$ and $$cm \equiv 1 \bmod p^k$$ then $$n \mapsto ncm+b p^k$$ is an isomorphism $$\Bbb{Z/p^k Z}^\times \to \{ g \in \Bbb{Z/mp^k Z}^\times, g \equiv 1 \bmod m\}$$ whose inverse is $$g \mapsto g\bmod p^k$$.
$$\Bbb{Z/p^k Z}^\times$$ is cyclic iff $$p$$ is an odd prime (the proof is that $$(1+p)^{p^l} \equiv 1+p^{l+1} \bmod p^{l+2}$$ and $$\Bbb{Z/p Z}$$ is a field thus all its elements are roots of $$x^{p}-x$$ but not of $$x^d-x$$ for any $$d < p$$)
• At first sight it is not obvious that this map is a homomorphism, but here is why: $b$ and $c$ can be chosen such that $bp^k+cm=1$, then $(icm+bp^k)(jcm+bp^k)=ijc^2m^2+b^2p^{2k}=ijcm(1-p^k)+bp^k(1-cm)=ijcm+bp^k$$. – Marc Bogaerts Jun 14 at 4:49 • @MarcBogaerts With$b,c$defined as I said the map is an homomorphism because$(cm) (bp^k) = 0 \bmod mp^k, (cm)^2 = cm \bmod mp^k, (bp^k)^2 = bp^k \bmod mp^k\$ – reuns Jun 14 at 21:59