Let $(\mathbb{Z}/n\mathbb{Z})^*$ denote the multiplicative group of units of $\mathbb{Z}/n\mathbb{Z}$. Is there a finite commutative group $G$ such that for all $n\geq 2$, there is no injective group homomorphism $f:G\to (\mathbb{Z}/n\mathbb{Z})^*$?
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asked Jan 31, 2018 at 16:53
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2 Answers
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No. By Dirichlet's theorem on primes in progressions, for each $n$ and $k$, there are at least $k$ primes $p_1,\dots,p_k$ congruent to $1$ modulo $n$, and so $$\left(\mathbb Z/ \left(\prod_{i=1}^k p_i\right) \mathbb Z\right)^\times$$ contains $(\mathbb Z/n)^k$ as a subgroup.
By the classification of finite commutative groups, every finite commutative group embeds into $(\mathbb Z/n)^k$ for some $n$ and $k$.
answered Jan 31, 2018 at 16:58
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Every finite abelian group $G$ is isomorphic to a product of the form $$\newcommand{\Z}{\mathbb Z}\Z/n_1\Z\times\dots\times\Z/n_k\Z.$$ Using Dirichlet's theorem on primes in arithmetic progressions (actually, a special case which can be proven elementarily using cyclotomic polynomials) pick primes $p_i$ such that $p_i\equiv 1\pmod n_i$ and let $n=p_1\dots p_k$. Then, by Chinese remainder theorem and cyclicity of $(\Z/p\Z)^*$, $$(\Z/n\Z)^*\cong(\Z/p_1\Z)^*\times\dots\times(\Z/p_k\Z)^*\cong\Z/(p_1-1)\Z\times\dots\times\Z/(p_k-1)\Z.$$ Since $n_i\mid p_i-1$, clearly $G$ embeds into $(\Z/n\Z)^*$.
