Let $n$ be a positive integer and let $d$ be some divisor of $n$. Consider the group of units modulo $n$, which we shall denote by $U(n)$. Likewise, denote the group of units modulo $d$ by $U(d)$.

Consider the homomorphism $f:\ U(n)\rightarrow U(d)$ given by reducing modulo $d$, $$f(u) = u \pmod{d}$$ I wish to show that this map is surjective. Can anyone supply a simple proof of this fact?

Alternatively, it is sufficient to prove that the set $$S = \left\{u\in U(n) \mid u\equiv 1 \pmod{d} \right\}$$ has cardinality $|S| = \phi(n)/\phi(d)$ where $\phi$ is the totient function.

  • $\begingroup$ For some reason I cannot view the mathjax rendered. If anyone can make sure the formatting is alright that would be great. $\endgroup$ – EuYu Sep 8 '13 at 2:13
  • $\begingroup$ Formatting looks good. $\endgroup$ – RghtHndSd Sep 8 '13 at 2:30

Let $u$ be a unit modulo $d$ and confuse it with the integer $0 < u < d$. Clearly if $(u, n) = 1$ then $u$ is in the image of the map. So suppose $(u,n) \neq 1$. Then it suffices to find $x \in \mathbb{Z}$ such that $(u + xd, n) = 1$. Indeed, if this can be done, then the map sends $u+xd \rightarrow u$ with $u+xd$ a unit in $\mathbb{Z}/n\mathbb{Z}$.

Claim: Taking $x$ to be a product of all primes dividing $n$ which do not divide $u$ suffices.

Proof: Let $p$ be a prime dividing $n$. If $p \mid u$, then $p$ does not divide $x$ and $p$ does not divide $d$ by definition. Hence $p \nmid u+dx$. Now suppose $p \nmid u$. Then $p \mid x$ by definition, so $p \nmid u + dx$.

Therefore every prime which divides $n$ does not divide $u+dx$, giving $(u+dx,n) = 1$.

  • $\begingroup$ Wonderful. Short and simple. Thank you. $\endgroup$ – EuYu Sep 8 '13 at 2:36

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