Prove/disprove are these Subgroups:
$\{(a,a):a\in G\}\subseteq G\times G$ where $G$ is a group
$H_1\times H_2\subseteq G_1\times G_2$ where $G_1,G_2$ groups and $H_1,H_2$ subgroups
- I am assuming the operation is addition as it is not mentioned.
let $a_1,a_2\in \{(a,a):a\in G\}$ that is $(a_1,a_1),(a_2,a_2)$ where $a_1,a_2\in G$.
We have to prove $(a_1,a_1)+(a_2^{-1},a_2^{-1})\in \{(a,a):a\in G\}$
$a_2\in G$ therefore $a_2^{-1}\in G$ so $(a_2^{-1},a_2^{-1})\in \{(a,a):a\in G\}$
$(a_1,a_1)+(a_2^{-1},a_2^{-1})=(a_1+a_2^{-1},a_1+a_2^{-1})$ we have $a_1+a_2^{-1}\in G$ so also $(a_1+a_2^{-1},a_1+a_2^{-1})\in \{(a,a):a\in G\}$
And therefore a subgroup
- a. $(e_1,e_2)\in H_1\times H_2$ so $H_1\times H_2\neq \emptyset$
b. $(a_1,a_2),(b_1,b_2)\in H_1\times H_2$ So $(a_1,a_2)+(b_1,b_2)=(a_1+b_1,a_2+b_2)$ $a_i+b_j\in H_i$ as $H_i$ is a subgroup so also do a_1+b_1,a_2+b_2)\in H_1\times H_2$
c. $(a_1,a_2)\in H_1\times H_2$ $a_1\in H_1$ and $a_2\in H_2$ which are subgroup so $a_1^{-1}\in H_1$ and $a_2^{-1}\in H_2$ so $(a_1^{-1},a_2^{-1})\in H_1\times H_2$
And therefore a subgroup
Are those proof valid?
Your proofs are valid, but I think they could be much more powerful if they were concisely written.
In the first proof, for example, you could just write: Clearly, the set $\{ (a,a):a∈G\}=\Delta (G)$ is not empty, because $(e,e)\in \Delta (G)$. Now, let $(a,a)\in \Delta (G)$ and $(b,b)^{-1}=(b^{-1},b^{-1})\in \Delta (G)$. As well, $(a,a)(b^{-1},b^{-1})=(ab^{-1},ab^{-1})\in \Delta (G)$, so that $\Delta (G)$ is a subgroup of $G\times G$.
For the second proof, a concise writting would be: If $e_1$ and $e_2$ are, respectively, the identities of $H_1$ and $H_2$, $H_1\times H_2$ is not empty, once that $(e_1,e_2)\in H_1\times H_2$. Now, let $(a,b)\in H_1\times H_2$ and $(c,d)^{-1}=(c^{-1},d^{-1})\in H_1\times H_2$. Therefore, $(a,b)(c^{-1},d^{-1})=(ac^{-1},bd^{-1})\in H_1\times H_2$, because $ac^{-1}\in H_1$ and $bd^{-1}\in H_2$. As a consequence, $H_1\times H_2$ is a subgroup of $G\times G$.