I have an exercise where I am tasked to prove that for $2$ subgroups $K < G$ and $J < H$ of $2$ groups $G,H$ the following is a subgroup: $$K \times J \subset G \times H$$
I believe I have done so, though I would appreciate some verification that my reasoning is correct.
Proof
The set $K \times J$ is defined as $\{ (k,j) \mid k \in K, j \in J\}$ where $K < G$ and $J< H$. We must show closure of products and inverses in $K \times J$ to show $K \times J < G \times H$.
- Let $(k,j)$, $(k',j') \in K \times J$. Then we have $$(k,j)(k',j') = (kk', jj') \in K \times J$$ Since $kk' \in K$ and $jj' \in J$
- Let $k^{-1} \in K$ and $j^{-1} \in J$ Then we have:$$(k,j)(k^{-1},j^{-1}) = (kk^{-1}, jj^{-1}) = (e,e)$$ $$(k^{-1},j^{-1})(k,j) = (k^{-1}k, j^{-1}j) = (e,e)$$ Since $kk^{-1} = k^{-1}k = e \in K$ and $jj^{-1} = j^{-1}j = e \in J$. Thus $(k^{-1},j^{-1}) \in K \times J$. Hence $(K \times J) < G \times H$. $\square$
Does this look alright?
You're almost there.
You need to show that the set is a nonempty subset in order to use the two-step subgroup test. Luckily, this is trivial: $(e_G, e_H)\in K\times J$ and $K\subseteq G$ & $J\subseteq H$ imply $K\times J\subseteq G\times H$.