A group $G$ is said to be Chernikov if it contains a normal subgroup N such that $G/N$ is finite and $N$ is direct product of finitely many Prufer groups.
The problem is the following:
If $G$ is a $2$-group and direct product of finitely many Chernikov ($2$-)groups, then is $G$ Chernikov?
Thank you in advance.
Let $G=G_1 \times \cdots \times G_r$, with $G_i$ $\check{C}$ernikov group for every $i=1, \ldots, r$. We want to show that $G$ is a $\check{C}$ernikov group by induction on $r$.
If $r=2$, let $G_1$ and $G_2$ be $\check{C}$ernikov groups and consider $G=G_1 \times G_2$. Since $G_1$ and $G_2$ are $\check{C}$ernikov groups, for $i \in \{1,2\}$ there exist $N_i \triangleleft G_i$ and $n_i \geq 0$ such that $G_i/N_i$ is finite and $N_i$ is direct product of $n_i$ quasicyclic groups. Consequently, if $N=\langle N_1, N_2 \rangle$, then $N$ is a normal subgroup of $G$, so we have to show that $G/N$ is finite and $N$ is direct product of $n=n_1 + n_2$ quasicyclic groups. Firstly, observe that $N_1 \cap N_2=1$ e che $N_1$ and $N_2$ are normal in $G$; thus, $N= N_1 \times N_2$, and $N$ is direct product of $n$ quasicyclic groups. Moreover, $N \triangleleft G$ because if $g_1g_2 \in G$ and $n_1 n_2 \in N$, since $xy=yx$ if $x \in G_1$ and $y \in G_2$, we have $$(n_1 n_2)^{g_1g_2}=(n_1^{g_1})^{g_2} (n_2^{ g_2})^{g_1}= n_1^{g_1} n_2^{g_2} \in N_1N_2= N.$$ Then, said $C$ the cartesian product of $G_1/N_1$ and $G_2/N_2$, the function which maps $g_1 g_2N \in G/N$ into $(g_1N, g_2N_2)\in C$ is a bijection. If $g_1g_2N=g_1'g_2'N$, then $g_2(g_1'g_2')^{-1} \in N$, i.e. $g_1g_1'^{-1}g_2g_2'^{-1} \in N_1N_2=N$. Thus from $g_1g_1'^{-1} \in N_1$ and $g_2g_2'^{-1} \in N_2$ it follows that $(g_1N_1, g_2 N_2)=(g_1'N_1, g_2'N_2)$. Finally, this function is bijective, being trivially surjective.
Now, let $r >2$ and suppose the assertion true for $r-1$ factors. Then, observing that $G=(G_1 \times \cdots \times G_{r-1}) \times G_r$, it's easy to show that $G$ is a $\check{C}$ernikov group.