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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.

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  • $\begingroup$ @reuns Sorry I mixed up in the title, corrected in the mean while $\endgroup$ – Marc Bogaerts Jun 12 at 16:16
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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 $d < p$)

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  • $\begingroup$ 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$$. $\endgroup$ – Marc Bogaerts Jun 14 at 4:49
  • $\begingroup$ @MarcBogaerts With $b,c$ defined as I said the map is an homomorphism because $(cm) (bp^k) = 0 \bmod mp^k, (cm)^2 = cm \bmod mp^k, (bp^k)^2 = bp^k \bmod mp^k$ $\endgroup$ – reuns Jun 14 at 21:59

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