# 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 $m$.

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 $$m$$?

• Without actually doing any work apart from just glancing at it, I wonder whether $G_1 \times \{e_2\}$ is always a subgroup of $G_1 \times G_2$, given that $\{e_2\}$ is the identity of $G_2$? Commented Dec 26, 2020 at 13:42
• @PrimeMover Yes, that's correct. Commented Dec 26, 2020 at 14:06

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 $$\lvert H\rvert=\lvert G_1\rvert=m$$.

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 $$G_1$$.

• Actually, you can't actually say $G_1$ is a subgroup of the product. It's not even a subset. All you can say is what you said in the previous sentence, which answers the question. Commented Dec 26, 2020 at 15:38
• @EthanBolker Formally you are right, but in practise you will simply identify $G_1$ as a subset of $G_1 \times G_2$ whenever convenient. Commented Dec 26, 2020 at 15:39
• True. But the "actually" suggests a literal interpretation. You might better say "Actually, that means you can often treat $G_1$ as if it were a subset of the product, much as you think of the real line as being the $x$-axis in the coordinate plane." Commented Dec 26, 2020 at 15:45
• I simply deleted the last line. Thanks for your comments. Commented Dec 26, 2020 at 18:49