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 $f:G\to (\mathbb{Z}/n\mathbb{Z})^\*$?



Subgroups of $(\mathbb{Z}/n\mathbb{Z})^*$
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 $k$.


