Subgroups of direct products

Question

Consider a group $G$ which is a direct product of two groups of coprime order:

$G = G_1 \times G_2$ with $|G_1|=n_1$, $|G_2|=n_2$ and $\textrm{gcd}(n_1, n_2)=1$.

Let $H \le G$. Is it true that $H=H_1 \times H_2$, where $H_1 \le G_1$ and $H_2 \le G_2$?

Answer 1

Because $H \le G$ we have that $|H|=k_1k_2$ where $k_i \bigm| n_i$ for $i=1,2$.

Define the mappings \begin{align*} \pi_i : \ \ \ \ \ \ \ \ \ \ G&\rightarrow G_i\\ (g_1, g_2)&\mapsto g_i \end{align*} for $i=1,2$. Both mappings are epimorphisms.

Consider $\pi_i(H)=H_i$. From the properties of epimorphisms we know that $H_i \le G_i$ and therefore $|H_i| \bigm| |G_i|=n_i$.

Consider the restrictions \begin{align*} \pi_i\big|_H : \ \ \ \ \ \ \ \ \ \ H&\rightarrow G_i\\ (g_1, g_2)&\mapsto g_1 \end{align*} Using the First Isomorphism Theorem it follows that $|H_i| \bigm| |H|=k_1k_2$.

It follows that $|H_i| \bigm| k_i$ and $|H_1 \times H_2| = |H_1| |H_2| \le k_1k_2 = |H| $.

However it is obvious that $H \subseteq H_1 \times H_2$, which implies that $|H| \le |H_1 \times H_2|$.

Therefore $|H|=|H_1 \times H_2|$ which demonstrates that $H=H_1 \times H_2$, as required.