[G,G] is normal closure of [S,S]

hbghlyj

If $G = \langle S \rangle$, then every element of $G$ can be written as a product of elements in $S$ and their inverses.
The commutator subgroup $[G,G]$ is the subgroup generated by all commutators:
$$[g,h] := ghg^{-1}h^{-1}, \quad \text{for } g,h \in G.$$
Define:$$[S,S] := \{ [s_1, s_2] \mid s_1, s_2 \in S \}.$$
Note that this is just a set, not a subgroup.

Question:

Is $[G,G]$ generated by the set of conjugates of $[S,S]$? That is,
$$
[G,G] = \langle \text{conjugates of } [s_1,s_2] \text{ for } s_1,s_2 \in S \rangle.
$$
More precisely: if $G = \langle S \rangle$, then:
$$
[G,G] = \langle [g,h] \mid g,h \in G \rangle = \langle x [s,t] x^{-1} \mid s,t \in S, x \in G \rangle.
$$

hbghlyj

What would go wrong if we only used $[S,S]$ without conjugates?
Example: $G = S_3$, $S = \{(12), (23)\}$
The commutators of the generators:
$$[(12), (23)] = (132).$$
So $[S,S] = \{(132)\}$.
But the commutator subgroup $[S_3, S_3]$ is $A_3 = \{ e, (123), (132) \}$.

Notice: We only got $(132)$ from $[S,S]$ directly. We are missing $(123)$.

How do we get (123)?
We take a conjugate:
$$
(12)(132)(12)^{-1} = (123).
$$
So, the other element of the commutator subgroup appears as a conjugate of a commutator of generators.

Moral of the story:
If you only take $[S,S]$, you don’t get a subgroup that contains all commutators of $G$.
You need all conjugates of $[S,S]$ to ensure you have a normal subgroup containing them — and $[G,G]$ is exactly the smallest normal subgroup containing all commutators.
$$
[G,G] = \langle g[s,t]g^{-1} \mid s,t \in S,\, g \in G\rangle
$$

hbghlyj

Let
\[
N := \big\langle g [s,t] g^{-1}
\bigm|s,t\in S,g\in G\big\rangle
\subseteq G.
\]

• Recall $[G,G]$ is the subgroup generated by all commutators $[g,h]$ with $g,h\in G$.
  To see each such $[g,h]\in N$, write
  \[
  g = s_{1}^{\varepsilon_1} s_{2}^{\varepsilon_2}\cdots s_{m}^{\varepsilon_m},
  \quad
  h = t_{1}^{\delta_1} t_{2}^{\delta_2}\cdots t_{n}^{\delta_n},
  \]
  with $s_i,t_j\in S$ and $\varepsilon_i,\delta_j=\pm1$.
  Then one uses the basic commutator identities
  \[
  [xy,z]
  = x [y,z] x^{-1}[x,z],
  \qquad
  [x,yz]
  = [x,y]y [x,z] y^{-1},
  \]
  repeatedly to express $[g,h]$ as a product of conjugates of $[s_i,t_j]$.
  Since each such conjugate lies in $N$, it follows that $[g,h]\in N$, and thus $[G,G]\subseteq N$.
• Note that for any $s,t\in S$ and any $g\in G$,
  \[
  g [s,t] g^{-1}
  =
  \bigl[g s g^{-1},g t g^{-1}\bigr],
  \]
  which is itself a commutator of elements of $G$.
  Hence each generator of $N$ lies in $[G,G]$, so $N\subseteq [G,G]$.