anon anon

2022-06-13

To prove that $\left[G,N\right]\subseteq N$ if and only if $N◁G$, we need to show the two directions of the implication separately.
First, let's assume that $\left[G,N\right]\subseteq N$. 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 $\left[G,N\right]\subseteq N$, it follows that for every $x\in G$ and $n\in N$, the commutator $\left[x,n\right]=xn{x}^{-1}{n}^{-1}$ is in $N$.
Now, consider an arbitrary element $n\in N$. We can rewrite $n$ as $\left[x,n\right]{x}^{-1}{n}^{-1}$ for some $x\in G$. Since $\left[x,n\right]$ is in $N$ and $N$ is a subgroup, we have $\left[x,n\right]{x}^{-1}{n}^{-1}\in N$.
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 $\left[G,N\right]\subseteq N$.
To show this, we need to demonstrate that for every $x\in G$ and $n\in N$, the commutator $\left[x,n\right]=xn{x}^{-1}{n}^{-1}$ 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 $xn{x}^{-1}{n}^{-1}=\left(xn{x}^{-1}\right){n}^{-1}\in N$.
Hence, we have shown that for every $x\in G$ and $n\in N$, $\left[x,n\right]=xn{x}^{-1}{n}^{-1}\in N$, which implies $\left[G,N\right]\subseteq N$.
By proving both directions, we have established the equivalence: $\left[G,N\right]\subseteq N$ if and only if $N$ is a normal subgroup of $G$.
To prove that $\left[G,N\right]\subseteq N$ if and only if $N◁G$, we will show the two directions of the implication separately.
First, assume that $\left[G,N\right]\subseteq N$. 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 $\left[G,N\right]\subseteq N$, it follows that for every $x\in G$ and $n\in N$, $\left[x,n\right]=xn{x}^{-1}{n}^{-1}\in N$.
Consider an arbitrary element $n\in N$. We can express $n$ as $\left[x,n\right]{x}^{-1}{n}^{-1}$ for some $x\in G$. Since $\left[x,n\right]\in N$ and $N$ is a subgroup, we have $\left[x,n\right]{x}^{-1}{n}^{-1}\in N$.
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 $\left[G,N\right]\subseteq N$. To show this, we need to demonstrate that for every $x\in G$ and $n\in N$, $\left[x,n\right]=xn{x}^{-1}{n}^{-1}\in N$.
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 $xn{x}^{-1}{n}^{-1}=\left(xn{x}^{-1}\right){n}^{-1}\in N$.
Hence, for every $x\in G$ and $n\in N$, $\left[x,n\right]=xn{x}^{-1}{n}^{-1}\in N$, which implies $\left[G,N\right]\subseteq N$.
By proving both directions, we have established the equivalence: $\left[G,N\right]\subseteq N$ if and only if $N$ is a normal subgroup of $G$.

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