on two decompositions of a finite group $G$ Let $G=A\times B\cong A\times C$. Then prove that there exists an isomorphisms $\gamma$ of $B$ to $C$ such that $Ab=Ab^\gamma$ for any $b\in B$.
we use $b^\gamma$ instead $(b)\gamma$. 
 A: For every $b \in B$, from the hypothesis,
$$ b = 1b \in A \times B = A \times C. $$
Hence, there exists unique $a_b \in A$ and $c_b \in C$ such that $b = a_b c_b$. Then define a map $\gamma \colon B \to C$ by $b^\gamma = c_b$.
The map $\gamma$ is a homomorphism:
For every $x, y \in B$,
$$\begin{align*}
(xy)^\gamma &= (a_x c_x a_y c_y)^\gamma \\
&= (a_x a_y c_x c_y)^\gamma & \text{($A$ and $C$ commute)} \\
&= c_x c_y = (a_x c_x)^\gamma (a_y c_y)^\gamma \\
&= x^\gamma y^\gamma.
\end{align*}$$
The homomorphism $\gamma$ is injection:
If $b^\gamma = 1$ then $b = a_b c_b = a_b b^\gamma = a_b$. Since $A$ and $B$ has trivial intersection, $b = 1$.
The homomorphism $\gamma$ is surjection:
For every $c \in C$, there exists $a \in A$ and $b \in B$ such that $c = ab$ since $c = 1c \in A\times C = A \times B$. Therefore $b^\gamma = (a^{-1}c)^\gamma = c$ as desired.
$Ab \subseteq Ab^\gamma$: For every $ab \in Ab$, 
$$ ab = a a_b c_b = a a_b b^\gamma \in Ab^\gamma. $$
$Ab^\gamma \subseteq Ab$: For every $ab^\gamma \in Ab^\gamma$,
$$ ab^\gamma = a c_b = a {a_b}^{-1} a_b c_b = a {a_b}^{-1} b \in Ab. $$
The proof completes.
A: As stated at the moment I am writing, the claim 

Let $G=A\times B\cong A\times C$. Then prove that there exists an isomorphisms $\gamma$ of $B$ to $C$ such that $Ab=Ab^\gamma$ for any $b\in B$.

is generally untrue.
Let $B_{i}$ be copies of a cyclic group $B$ of order $4$, and $C_{i}$ copies of a cyclic group $C$ of order $2$. Consider
\begin{equation*}
A = \prod_{i=1}^{\infty} B_{i} \times \prod_{i=1}^{\infty} C_{i}.
\end{equation*}
Then clearly
\begin{equation*}
A \times B \cong A \cong A \times C
\end{equation*}
while of course $B \not\cong C$.
