To prove that $[G,N]\subseteq N$ if and only if $N\u25c1G$, we need to show the two directions of the implication separately.

First, let's assume that $[G,N]\subseteq N$. We want to prove that $N$ is a normal subgroup of $G$.

To show that $N$ is normal, we need to demonstrate that for every $g$ in $G$, $gN{g}^{-1}\subseteq N$.

Let $g$ be an arbitrary element in $G$. Since $[G,N]\subseteq N$, it follows that for every $x\in G$ and $n\in N$, the commutator $[x,n]=xn{x}^{-1}{n}^{-1}$ is in $N$.

Now, consider an arbitrary element $n\in N$. We can rewrite $n$ as $[x,n]{x}^{-1}{n}^{-1}$ for some $x\in G$. Since $[x,n]$ is in $N$ and $N$ is a subgroup, we have $[x,n]{x}^{-1}{n}^{-1}\in N$.

Therefore, we have shown that for every $g\in G$, $gN{g}^{-1}\subseteq N$, which proves that $N$ is a normal subgroup of $G$.

Next, let's assume that $N$ is a normal subgroup of $G$. We want to prove that $[G,N]\subseteq N$.

To show this, we need to demonstrate that for every $x\in G$ and $n\in N$, the commutator $[x,n]=xn{x}^{-1}{n}^{-1}$ is in $N$.

Since $N$ is a normal subgroup of $G$, we have $xn{x}^{-1}\in N$ for every $x\in G$ and $n\in N$.

Let's consider an arbitrary $x\in G$ and $n\in N$. Since $N$ is a subgroup, ${n}^{-1}\in N$ as well. Therefore, we have $xn{x}^{-1}{n}^{-1}=(xn{x}^{-1}){n}^{-1}\in N$.

Hence, we have shown that for every $x\in G$ and $n\in N$, $[x,n]=xn{x}^{-1}{n}^{-1}\in N$, which implies $[G,N]\subseteq N$.

By proving both directions, we have established the equivalence: $[G,N]\subseteq N$ if and only if $N$ is a normal subgroup of $G$.

To prove that $[G,N]\subseteq N$ if and only if $N\u25c1G$, we will show the two directions of the implication separately.

First, assume that $[G,N]\subseteq N$. We want to prove that $N$ is a normal subgroup of $G$. To do this, we need to show that for every $g\in G$, $gN{g}^{-1}\subseteq N$.

Let $g$ be an arbitrary element in $G$. Since $[G,N]\subseteq N$, it follows that for every $x\in G$ and $n\in N$, $[x,n]=xn{x}^{-1}{n}^{-1}\in N$.

Consider an arbitrary element $n\in N$. We can express $n$ as $[x,n]{x}^{-1}{n}^{-1}$ for some $x\in G$. Since $[x,n]\in N$ and $N$ is a subgroup, we have $[x,n]{x}^{-1}{n}^{-1}\in N$.

Therefore, for every $g\in G$, $gN{g}^{-1}\subseteq N$, which proves that $N$ is a normal subgroup of $G$.

Next, assume that $N$ is a normal subgroup of $G$. We want to prove that $[G,N]\subseteq N$. To show this, we need to demonstrate that for every $x\in G$ and $n\in N$, $[x,n]=xn{x}^{-1}{n}^{-1}\in N$.

Since $N$ is a normal subgroup of $G$, we have $xn{x}^{-1}\in N$ for every $x\in G$ and $n\in N$.

Consider an arbitrary $x\in G$ and $n\in N$. Since $N$ is a subgroup, ${n}^{-1}\in N$ as well. Therefore, we have $xn{x}^{-1}{n}^{-1}=(xn{x}^{-1}){n}^{-1}\in N$.

Hence, for every $x\in G$ and $n\in N$, $[x,n]=xn{x}^{-1}{n}^{-1}\in N$, which implies $[G,N]\subseteq N$.

By proving both directions, we have established the equivalence: $[G,N]\subseteq N$ if and only if $N$ is a normal subgroup of $G$.