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
$G_1times G_2={(bar0,bar0),(bar0,bar1),(bar0,bar2),(bar1,bar0),(bar1,bar1),(bar1,bar2)}$
I see that $o(G_1times G_2)=o(G_1)cdot o(G_2)=mcdot n=6.$ 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 $o(G_1times G_2)=o(G_1)cdot o(G_2)$ and there will always be a subgroup of order $m$?

Shaun · Accepted Answer

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

QuantumSpace · Answer

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 $G_1$.

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 $m$.

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