Isomorphism of Group Products Let $G$ be a group, $A = G \times G$. In $A$, Let $T = \{(g, g)|g \in G\}$. Prove that $T$ is isomorphic to $G$.
I don't know how to continue this problem. $A$ is abelian. Therefore, $G \times G$ is abelian. $T$ is a subgroup of $G$.
 A: Define $f: T\to G$ by $f(g,g)=g$.
Then $f((g_1,g_1)(g_2,g_2))=f((g_1g_2,g_1g_2))=g_1g_2=f((g_1,g_1))f((g_2,g_2))$ So $f$ is a group homomorphism.
To show oneone ; $f((g_1,g_1))=f((g_2,g_2))\Rightarrow g_1=g_2\Rightarrow (g_1,g_1)=(g_2,g_2)$ 
To show onto: $\forall g\in G$ we have $(g,g)\in T$ and $f((g,g))=g$
So $f$ is an isomorphism.
A: Outline: Define $\varphi: T\to G$ by $\varphi(g,g)=g$. Show that $\varphi$ is an isomorphism. Checking that $\varphi$ is an isomorphism will not be difficult.
Note that $T$ is not a subgroup of $G$, it is a subgroup of $G\times G$. There is no need to assume that $G$ is Abelian. The result holds for all groups.   
A: As others have noted it is not hard to show that $\phi: G \to T$, $g \mapsto (g,g)$ is a homomorphism. However I'd like to explicitly point out that there is an easy way to show this is bijective, when asked if something is bijective it is sometimes easier to just show it has an inverse function: Define $\psi: T\to G$ by sending $(g,g)$ to $g$. Alternative let $\psi = \pi \circ i$ where $i: T \to G \times G$ is inclusion and $\pi:G \times G \to G$ is a projection onto one coordinate. Then $\psi \circ \phi = \text{id}_G$, $\phi\circ\psi = \text{id}_T$.
