Commutator subgroup $G'$ is a characteristic subgroup of $G$

Question

For any group $G$, prove that the commutator subgroup $G'$ is a characteristic subgroup of $G$.

Let $U=\{xyx^{-1}y^{-1}|x, y \in G\}$. Now $G'$ is the smallest subgroup of $G$ which contains $U$. We need to show that $T(G') \subset G'$ for all automorphisms $T$ of $G$.

If every element of $G'$ is of the form $g'=\{xyx^{-1}y^{-1}\}$, then $T(xyx^{-1}y^{-1 })=T(x)T(y)T(x^{-1})T(y^{-1})=T(x)T(y){T(x)}^{-1}{T(y)}^{-1}$. Since $T$ is an automorphism $T(g) \in G$ for all $g\in G$. Hence $T(x)T(y){T(x)}^{-1}{T(y)}^{-1}\in G$. But this is of the form of an element in $U$. Hence $T(g')\in U$. So $T(g') \in G'$.

Is this the end of it?? I mean is every element in $G'$ of the form mentioned above??

Elements of $G'$ are not necessarily commutators, but you can write them as a product of commutators. — Seirios, Apr 25 '14 at 05:19
You can show the following general fact: Let $G$ be a group and $H$ be a subgroup generating by a subset $S$. Then $H= {s_1^{\pm 1} \cdots s_n^{\pm 1} \mid n \geq 0 }$. — Seirios, Apr 25 '14 at 05:28
@Seirios: I think you meant to write $H = {s^\pm_1\cdots s^\pm_n ,, | ,, n \ge 0, s_1,\ldots,s_n \in S}$. — Lee Mosher, Apr 25 '14 at 14:08

Lee Mosher · Accepted Answer · 2021-12-25T18:20:09.520

The point is not that every element of $G'$ is an element of $U$; this is false, as noted in the comments. In general if $G$ is a group and $U$ is a subset, the "smallest subgroup of $G$ containing $U$" need not be equal to $U$.

The point instead is that the set $U$ is preserved by every automorphism $T$, in other words $T(U)=U$. For instance to prove that $T(U) \subset U$, take an element of $T(U)$ of the form $T(g)$ where $g = x y x^{-1} y^{-1} \in U$. Then $T(g) = T(x y x^{-1} y^{-1}) = T(x) T(y) T(x)^{-1} T(y)^{-1} \in U$. The reverse inclusion $U \subset T(U)$ is equivalent to $T^{-1}(U) \subset U$ which is proved similarly.

To say that $G'$ is the smallest subgroup of $G$ containing $U$ means that $G'$ is the intersection of all subgroups of $G$ containing $U$. From this it follows that $T(G')$ is the intersection of all subgroups of $G$ containing $T(U)$, i.e. $T(G')$ is the smallest subgroup of $G$ containing $T(U)$. To see why this is true, notice that like any permutation of any set, the automorphism $T$ also permutes the set of subsets of $G$. The latter permutation preserves the inclusion relation and the intersection operation: for any subsets $A \subset B$ of $G$ we have $T(A) \subset T(B)$; and for any collection of subsets $\{A_i\}$ we have $T\bigl( \cap_i A_i \bigr) = \cap_i T(A_i)$. Also, because $T$ is a group automorphism, a subset $A \subset G$ is a subgroup if and only if $T(A) \subset G$ is a subgroup. Since $G'$ is the intersection of all subgroups that contain $U$, it follows that $T(G')$ is the intersection of all subgroups that contain $T(U)$.

Since $T(U)=U$, it follows that $T(G')$ is the smallest subgroup of $G$ containing $U$, and so $T(G')=G'$.

How do we know $T(G')$ is the smallest subgroup of $G$ containing $T(U)$? — user5826, Dec 05 '17 at 18:11
All we know is that $G'$ is the smallest subgroup containing $U$. — user5826, Dec 06 '17 at 01:50
I have added a paragraph to explain the set theoretic details behind the implication that is questioned in these comments. — Lee Mosher, Dec 25 '21 at 18:22
@LeeMosher If $T$ is merely a homomorphism then still we have $T(G')\subset G'$? — Meet Patel, Apr 30 '24 at 13:20

Commutator subgroup $G'$ is a characteristic subgroup of $G$

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