$G\times G\cong H\times H\Longrightarrow G\cong H$ for $G$ ACC and DCC

Question

I want to prove that, if $G$ satisfies ACC and DCC on normal subgroups, then $G\times G\cong H\times H$ implies $G\cong H$.

I observed that if we can prove that $G\times G$ satisfies ACC and DCC the conclusion will follow by Krull-Schmidt. So I tried to see if it's true

Let $\pi_1:G\times G\to G$ and $\pi_2:G\times G\to G$ be projection on the first and second summand. Then any $\{e\}\le N_1\le N_2\le\cdots$. Then $\pi_1(N_k)$ and $\pi_2(N_k)$ both stabilize, but I'm not sure if I can conclude that $N_k$ also stabilizes.

1. Is what I am trying to prove in the second paragraph even correct?
2. If not, how do I prove the theorem in the first paragraph?

Answer 1

It is true that if $G$ satisfies ACC and DCC then $G \times G$ also does:

Let $N_1 \leq N_2 \leq \cdots$ in $G \times G$. Then let $K_i=\pi_1(N_i)$ and $L_i=(\{e\} \times G) \cap N_i$. Both stabilize i.e. there is an $n\geq 1$ such that for all $m \geq n$ we have $L_m=L_n$ and $K_m=K_n$. We want to show that for all $m \geq n$ we also have $N_m=N_n$. Let $x \in N_m$ then $\pi_1(x) \in K_m=K_n$. Therefore, there is an $x' \in N_n$ such that $\pi_1(x')=\pi_1(x)$ which implies $x'-x \in \ker(\pi_1)= \{e\} \times G$. We have $x,x' \in N_m$ thus $x'-x \in L_m=L_n$. Now $L_n \subseteq N_n$ and $x' \in N_n$ imply $x \in N_n$.

Analogously, we can proceed for DCC.