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Let $G$ be a finite abelian group.

How do I show that $G$ can be embedded in $(\mathbb{Z}_n)^*$ for some n?

Also, that there exists a subgroup $H$ such that $G$ is isomorphic to $(\mathbb{Z}_n)^*/H$

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As to your second question: if $G$ is an abelian group, then for every subgroup $K$ of $G$ there is a subgroup $H$ of $G$ such that $G/H\cong K$, and conversely (that is, every subgroup is isomorphic to a quotient, and every quotient is isomorphic to a subgroup). – Arturo Magidin Apr 22 '12 at 21:08
up vote 6 down vote accepted

We can write $G=\prod_{i=1}^tC_{m_i}$ as direct product of cyclic groups.

One way of showing this is to first find prime numbers $p_i$ such that $p_i\equiv 1\pmod{m_i}$. Furthermore, we require that the primes $p_i$ should be distinct. This is possible, because by Dirichlet's theorem of primes in an arithmetic progression there are infinitely many primes $p_i$ satisfying the above congruence for a given $m_i$.

The group $\mathbb{Z}_{p_i}^*$ is cyclic of order $p_i-1$, so it has a subgroup isomorphic to $C_{m_i}$. It also has a cyclic subgroup $H_i$ of order $(p_i-1)/m_i$, and $\mathbb{Z}_{p_i}^*/H_i\cong C_{m_i}$. Let $n=\prod_{i=1}^tp_i$. By the Chinese Remainder Theorem $\mathbb{Z}_n^*\cong\prod_{i=1}^t\mathbb{Z}_{p_i}^*$.

The rest is easy.

Note: AFAICT using the fact that arithmetic progressions contain infinitely many primes is unavoidable here. For example, if there were only finitely many primes congruent to $1$ modulo a given prime $q$, then we would run into problems. Let $p_1,p_2,\ldots,p_k$ be those primes, and let $n=q^2p_1p_2\cdots p_k$. Then we can embed $C_q^{k+1}$ into $\mathbb{Z}_n^*$, but $C_q^{k+2}$ would not be a subgroup of $\mathbb{Z}_n^*$ for any $n$. The reason is that if $n=q^mp_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}\prod_j \ell_j^{b_j}$ is the prime factorization of $n$, and here the primes $\ell_j\not\equiv1\pmod{q}$, we easily see that the $q$-torsion subgroup of $\mathbb{Z}_n^*$ has size $<q^{k+2}$.

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