Exercise 3.1.37 ($(A imes B) /(A imes \{1\}) \simeq B$)

Question

Let $A$ and $B$ be groups. Show that $\{(a,1) \mid a \in A\}$ is a normal subgroup of $A 	imes B$ and the quotient of $A 	imes B$ by this subgroup is isomorphic to $B$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Consider the set
$$H = A 	imes \{1\} = \{(a,1) \mid a \in A\}.$$
(In other words, $H = \{(a,b) \in A 	imes B \mid b = 1\}$).

Since $(1,1) \in H$, $H 
e \varnothing$, and $H = A 	imes \{1\} \subseteq A 	imes B$. If $h,k \in H$, there are elements $a, b \in A$ such that $h = (a,1)$ and $k = (b,1)$. Then
$$hk^{-1} = (a,1)(b^{-1}, 1) = (ab^{-1}, 1) \in H,$$
so $H$ is a subgroup of $A 	imes B$.

Moreover, if $g = (u,v) \in A 	imes B$, and $h = (a,1) \in H$, then
$$g h g^{-1} = (u,v) (a,1) (u^{-1}, v^{-1}) = (uau^{-1}, uvv^{-1}) = (uau^{-1},1) \in H,$$
so
$$A 	imes \{1\} \unlhd A 	imes B.$$


Consider the map
$$\varphi\
\left\{
\begin{array}{ccl}
A 	imes B & 	o & B\
(a,b) & \mapsto & b.
\end{array}
ight.
$$
Then 
\begin{itemize}
\item $\varphi$ is a homomorphism: If $u = (a,b) \in A 	imes B$ and $v = (c,d) \in A 	imes B$, then
$$\varphi(uv) = \varphi((a,b)(c,d)) = \varphi(ac, bd) = bd = \varphi((a,b)) \varphi((c,d)) = \varphi(u) \varphi(v).$$

\item $\varphi$ is surjective: If $b$ is any element of $B$, then $\varphi(1,b) = b$.

\item $\ker(\varphi) = A 	imes\{1\} = H$: If $u =(a,b) \in A 	imes B$, then
$$u \in \ker(\varphi) \iff b = 1 \iff u \in A 	imes \{1\}.$$

\end{itemize}
By the First Isomorphism Theorem, $ (A	imes B)/ \ker(\varphi) \simeq \varphi(A 	imes B) = B$, so
$$(A 	imes B) /(A 	imes \{1\}) \simeq B.$$
\end{proof}

Exercise 3.1.37 ($(A \times B) /(A \times \{1\}) \simeq B$)

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Exercise 3.1.37 ($(A \times B) /(A \times \{1\}) \simeq B$)

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