Given the diagonal $H = \{(x, x): x\in G\}$, show that (i) $H$ is a subgroup of $G\times G = G^2$ and; (ii) $H \cong G$
My attempt:
Proof. (i) $G$ is a group, so an identity element $e\in G$ exists. By extension, $(e, e)\in H$. Let $x = (a, a), y = (b, b)$ for any two elements $a, b \in G$. Then $a, b \in H$. $G$ contains an inverse element, so $a^{-1}, b^{-1} \in G$. Then $x^{-1}, y^{-1}$ eixst on $H$, too. Therefore, $$y^{-1} = (b^{-1}, b^{-1}) \\ \implies xy^{-1} = (ab^{-1}, ab^{-1})\in H$$ $H$ is a subgroup of $G^2$.
(ii) Let $\phi: H \rightarrow G$ be defined by $\phi(a, a) = b$ where $(a, a)\in H$ and $b\in G$. If $\phi(a, a) = \phi (a', a')$ then necessarily we must have $a = a'$. $\phi$ is injective. The mapping $\rho: G\rightarrow H $ determined by $\rho(b) = (a, a)$ is injective since $(a, a) = (b, b)$ if and only if $a = b$. Hence, a bijection $H\rightarrow G$ exists. Finally, consider the function $\mu(a) = (a, a)$. Then \begin{align*}\mu(ab) &= (ab, ab) \\ &= (a, a)(b, b) \\ &= \mu(a)\mu(b)\end{align*} gives us a homomorphism.
Therefore, $H$ is isomorphic to $G$. $\blacksquare$
Is there any flaw with my proof? Please let me know.
If you know that the image of a group under a group homomorphism is a subgroup of the codomain group, then everything can be done at once by considering
$$\begin{array}{l|rcl} \varphi: & G& \longrightarrow & H \subseteq G \times G\\ & x & \longmapsto & (x,x) \end{array}$$
It is easy to prove that $\varphi$ is a group homomorphism, one-to-one as $\varphi^{-1}[\{(e,e)\}]=\{e\}$ and onto on $H$.
Note: this however depends on where you’re in your groups course. In particular if you already covered or not group homomorphism theorems and product of groups.