Exercise 3.1.38 ($(A imes A)/D \simeq A$)

Question

Let $A$ be an abelian group and let $D$ be the (diagonal) subgroup $\{(a,a) \mid a \in A\}$ of $A 	imes A$. Prove that $D$ is a normal subgroup of $A 	imes A$ and $(A 	imes A)/D \simeq A$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $A$ be an abelian group (in multiplicative notation) and let
$$D = \{(a,a) \mid a \in A\}.$$
(In other words, $D = \{(a,b) \in A 	imes B \mid a = b\}$.)

Then $D \leq A 	imes A$, and since $A$ is abelian, then $A 	imes A$ is abelian, so every subgroup is normal, in particular $D \unlhd A 	imes A$.
Consider the map
$$\varphi\
\left\{
\begin{array}{ccl}
A 	imes A & 	o & A\
(a,b) & \mapsto & ab^{-1}.
\end{array}
ight.
$$
Then
\begin{itemize}
\item $\varphi$ is a homomorphism: Since $A$ is abelian, for all $u = (a,b) \in A 	imes A$ and all $v = (c,d) \in A 	imes A$,
$$\varphi(u) \varphi(v) = \varphi(a,b) \varphi(c,d) = ab^{-1} cd^{-1}  ac (bd)^{-1} = \varphi(ac, bd)= \varphi((a,b)(c,d)) = \varphi(uv).$$

\item $\varphi$ is surjective: Let $a$ be any element of $A$. Then $a = \varphi(a,1)$, where $(a,1) \in A 	imes A$, so $\varphi$ is surjective, and $\varphi(A 	imes A) = A$.

\item $\ker(\varphi) = D$: If $(a,b) \in A 	imes A$, then
$$(a,b) \in \ker(\varphi) \iff ab^{-1} = 1 \iff a = b \iff (a,b) \in D.$$
\end{itemize}
By the First Isomorphism Theorem, $(A	imes A) / \ker(\varphi) \simeq \varphi(A 	imes A)$, so
$$(A 	imes A)/D \simeq A.$$
\end{proof}

Exercise 3.1.38 ($(A \times A)/D \simeq A$)

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Exercise 3.1.38 ($(A \times A)/D \simeq A$)

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