Dummit and Foote, Section 3.1, Problem 29 states:
"Let $N$ be a finite subset of $G$ and suppose $G = \langle T \rangle$ and $N = \langle S \rangle$ for some subsets $S$ and $T$ of $G$. Prove that $N$ is normal in $G$ if and only if $tSt^{-1} \subset N$ for all $t \in T$."
The authors emphasize the "finite". Furthermore, the previous two questions (27 and 28) require $N$ to be finite for the statement to hold. My problem is that I seem to have reasoned out a "proof" that does not use the fact that $N$ is finite anywhere.
Here's my proposed proof:
$\rightarrow$) Suppose $N$ is normal in $G$, $t \in T$ and $x \in tSt^{-1}$. By definition, $x = tst^{-1}$ for some $s \in S$. Since $N = \langle S \rangle$, $s \in N$, and $x \in tNt^{-1}$. Since $N$ is normal in $G$ and $t \in G$, $tNt^{-1} \subset N$. Thus, $x \in N$. It follows that $tSt^{-1} \subset N$.
$\leftarrow$) Suppose $tSt^{-1} \subset N$ for all $t \in T$, and let $g \in G$. Since $G = \langle T \rangle$, $g = t_1^{\alpha_1}…t_n^{\alpha_n}$, where $t_i \in T$ and $\alpha_i \in \mathbb{Z}$ for $i \in \{1,…,n\}$. Then, for some $n \in N$, $$x = gng^{-1} = (t_1…t_{n-1}^{\alpha_{n-1}})t_n^{\alpha_n}nt_n^{-\alpha_n}(t_{n-1}^{-\alpha_{n-1}}…t_1^{-\alpha_1}).$$
Since $N = \langle S \rangle$, $n = s_1^{\beta_1}…s_m^{\beta_m}$ where $s_i \in S$ and $\beta_i \in \mathbb{Z}$ for $i \in \{1,…,m\}$. From above, $$x = (t_1…t_{n-1}^{\alpha_{n-1}})t_n^{\alpha_n}(s_1^{\beta_1}…s_m^{\beta_m})t_n^{-\alpha_n}(t_{n-1}^{-\alpha_{n-1}}…t_1^{-\alpha_1}).$$
Examining the middle of this term, we can use properties of conjugates to simplify:
$$t_n^{\alpha_n}(s_1^{\beta_1}…s_m^{\beta_m})t_n^{-\alpha_n} = (t_n (s_1^{\beta_1}…s_m^{\beta_m}) t_n^{-1})^{\alpha_n} = ((t_ns_1t_n^{-1})^{\beta_1}…(t_ns_mt_n^{-1})^{\beta_m})^{\alpha_n}.$$ Since $tSt^{-1} \subset N$ for all $t \in T$, and since $N$ is a subgroup and closed under multiplication, all terms above are in $N$, hence the entire term is in $N$ and $x \in (t_1…t_{n-1}^{\alpha_{n-1}})N(t_{n-1}^{-\alpha_{n-1}}…t_1^{-\alpha_1})$. Repeating this process $n$ times will eventually give us $x \in N$, so $gNg^{-1} \subset N$ and by Theorem 6(5) of the textbook $N$ is a normal subgroup.
My concern: I do not see where I use the fact that $N$ is a finite subgroup in my proposed proof. Can somebody check my proof for flaws?
Best Answer
The issue is that if $G$ is infinite then you cannot write an element $g\in G$ as a product of elements from $T$, but instead $g$ is a product of elements of $T$ and $T^{-1}$. For example, $(\mathbb{Z},+)=\langle1\rangle$ but no negative integer is a sum of $1$s.
So the issue is here:
You need to take $\alpha_i\in\mathbb{Z}$, but doing so violates the assumptions you are working under.
(The result would hold if $T$ was closed under inverses.)