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Prove/disprove are these Subgroups:

  1. $\{(a,a):a\in G\}\subseteq G\times G$ where $G$ is a group

  2. $H_1\times H_2\subseteq G_1\times G_2$ where $G_1,G_2$ groups and $H_1,H_2$ subgroups

  1. 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

  1. 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?

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    $\begingroup$ Side note: it does not really matter whether you use addition or multiplication to denote the group operation (multiplication is more common except for abelian groups) but you should stay consistent. If you want to use addition, write the inverse of $a$ as $-a$; if you want to use multiplication, write the inverse of $a$ as $a^{-1}$. $\endgroup$ – angryavian Feb 13 '18 at 17:35
  • $\begingroup$ I see, what I meant is that in the qeustion I was not given what is the operation, and it crucial as a set with the ordinary addition can be a group whereas the same set with multiplication may not be a group. But thanks for the comment I will try to be consistent from now on $\endgroup$ – gbox Feb 13 '18 at 17:41
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    $\begingroup$ The ingredients seem to be all there. But I must say it is poorly written, both from the phrase construction point of view and from the mathematical editing one. You should improve those skills. $\endgroup$ – amrsa Feb 13 '18 at 18:23
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    $\begingroup$ For 1, you also need to show the set $\{(a,a):a\in G\}$ is non-empty. $\endgroup$ – Delong Feb 13 '18 at 19:23
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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 $G\times G$.

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