Prove that $G$ has a subgroup isomorphic to $G/H$.

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.

• Since H is normal in G, wouldn't the coset representatives from G/H be a subgroup of G isomorphic to G/H? – Laars Helenius Nov 13 '14 at 4:47
• So the isomorphism would be $\varphi(gH)=g$? – Laars Helenius Nov 13 '14 at 4:49
• @LaarsHelenius: No it cannot be that simple. Your idea already fails for $G=\Bbb Z/4\Bbb Z$ and $|H|=2$. – Marc van Leeuwen Nov 14 '14 at 12:43

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 $G/H$.
Hint: If you know that the quotient of a direct product is isomorphic to the direct product of quotients, then you can find the structure of $G/H$.
That is, if $G = G_1 \times \cdots \times G_k$ and $H = H_1 \times \cdots \times H_k$ such that $H_i \unlhd G_i$ $\forall i$, then $$G/H = (G_1 \times \cdots \times G_k) / (H_1 \times \cdots \times H_k) = (G_1/H_1) \times \cdots \times (G_k/H_k).$$
You need to write $G$ and $H$ so that they have the same amount of terms in the products. That is, write the trivial terms too.
• The problem is that is is not clear that you can have compatible direct product decompositions for $G$ and $H$. – Marc van Leeuwen Nov 14 '14 at 12:47