If given two groups $G_1,G_2$ of order $m$ and $n$ respectively then the direct product $G_{1}\times G_{2}$ has a subgroup of order $m$.

Question

I'm trying to understand if this statement is false or true, I have tried understanding by an example.

For example if $G_1=(\{{\bar0},\bar1\},+_2),G_3=(\{{\bar0},\bar1,\bar2\},+_3)$ then the direct product of these two groups will be

$G_1\times G_2=\{(\bar0,\bar0),(\bar0,\bar1),(\bar0,\bar2),(\bar1,\bar0),(\bar1,\bar1),(\bar1,\bar2)\}$

I see that $o(G_1\times G_2)=o(G_1)\cdot o(G_2)=m\cdot n=6.$ Can I simply say that I take the group $\{{(\bar0,\bar0),(\bar0,\bar1)}\}$ under the addition modulo 2 and 3 respectively and the statement is true for this case ?

Also if this statement is true shouldn't it mean that it will always be true considering that $o(G_1\times G_2)=o(G_1)\cdot o(G_2)$ and there will always be a subgroup of order $m$?

Answer 1

Hint: Consider $$H=\{(g, e_{G_2})\mid g\in G_1\}.$$

Use the one-step subgroup test. Clearly $H\subseteq G_1\times G_2$. Since $G_1$ is a group, its identity $e_{G_1}\in G_1$, so $(e_{G_1}, e_{G_2})\in H\neq \varnothing$. Let $x=(g_1, e_{G_2})$, $y=(g_2, e_{G_2})\in H.$ Then $$\begin{align}xy^{-1}&=(g_1, e_{G_2})(g_2, e_{G_2})^{-1}\\ &=(g_1, e_{G_2})(g_2^{-1}, e_{G_2})\\ &=(g_1g_2^{-1}, e_{G_2})\\ &\in H\end{align}$$ since $g_1g_2^{-1}\in G_1$ as $G_1$ is a group. Hence $H\le G_1\times G_2$. But clearly $\lvert H\rvert=\lvert G_1\rvert=m$.

Answer 2

Here is another way to see that $G_1 \cong G_1 \times \{e_{G_2}\}$ is a subgroup.

Note that the map $$\phi: G_1 \to G_1 \times G_2: g \mapsto (g, , e_{G_2})$$ is a group morphism. Hence, $\text{Im}(\phi) = G_1 \times \{e_{G_2}\}$ is a subgroup of $G_1 \times G_2$ and this is the subgroup you are looking for.

Note moreover that $\phi$ is an injection, so one can even say that $G_1 \times G_2$ contains a subgroup isomorphic to $G_1$.