Let $G$ be a finite abelian group of order $n$ and let $H$ be a subgroup of $G$ of order $m$. Show that $G$ has a subgroup isomorphic to $G/H$.
Here are my thoughts:
Define $\mu_n := \{z \in \mathbb{C}|z^n=1\}$ and $\mu:=\bigcup_{n=1}^{\infty}\mu_n$.
Since a finite abelian group is self-dual, we have $G/H \cong \operatorname{Hom}(G/H,\mu)$.
By the structure theorem for finite abelian groups, we have $$G \cong \mathbb{Z}_{n_1}\times \cdot \cdot \cdot \times \mathbb{Z}_{n_s}$$ where $n_{i+1}|n_i$ for $1 \leq i \leq s-1$ and $n_1\cdot \cdot \cdot n_s=n$, and $n_i \geq 2$ for all $i$.
Also, we have $$H \cong \mathbb{Z}_{m_1}\times \cdot \cdot \cdot \times \mathbb{Z}_{m_l}$$ where $m_{i+1}|m_i$ for $1 \leq i\leq l-1$ and $m_1 \cdot \cdot \cdot m_l=m$, and $m_i \geq 2$ for all $i$.
I want to somehow write $G/H$ as a product of cyclic subgroups of the $\mathbb{Z}_{n_i}$, for now denote this product by $A_1\times \cdot \cdot \cdot \times A_s$, where $A_i \leq \mathbb{Z}_{n_i}$, so that then I can write $$\operatorname{Hom}(G/H,\mu) \cong \operatorname{Hom}(A_1, \mu)\times \cdot \cdot \cdot \times \operatorname{Hom}(A_s, \mu).$$
Then, since $\operatorname{Hom}(A_i,\mu) \leq \operatorname{Hom}(\mathbb{Z}_{n_i},\mu)$ and (by an already known result) $\operatorname{Hom}(\mathbb{Z}_{n_i},\mu) \cong \mathbb{Z}_{n_i}$, I will have $$\operatorname{Hom}(G/H, \mu) \leq \mathbb{Z}_{n_1}\times \cdot \cdot \cdot \times \mathbb{Z}_{n_s}. $$
My problem is I cannot see how to write $G/H$ as such a product or if it's even possible. I appreciate any guidance and suggestions. Thanks.
Best Answer
I think the self-duality mentioned in the question makes this (otherwise rather difficult) question easy. In $\widehat G=\def\Hom{\operatorname{Hom}}\Hom(G,\mu)$ consider $\{f\in\widehat G\,\mid H\subseteq\ker(f)\,\}$, a subgroup that is canonically isomorphic to $\Hom(G/H,\mu)=\widehat{G/H}$, and therefore (non-canonically) to $G/H$. Under the (non-canonical) isomorphism $\widehat G\to G$, it maps to a subgroup of $G$ that is isomorphic to $G/H$.