Checking whether the group generated by a set is normal As part of a homework problem, we proved that we can test whether an element $g$ of a finite group $G$ is in the normalizer of a cyclic subgroup $H=\langle x\rangle$ by conjugating just the generator $x$ by $g$,
$$gxg^{-1}=x^{a} \iff g\in N_{G}(H)$$
for some $a\in\mathbb{Z}$. Namely, this means we can test a cyclic group for normality by merely conjugating the single generator and seeing if the result is a power of $x$: $$H\trianglelefteq G \iff  \forall g\in G, gxg^{-1}=x^{a}$$
for some $a\in \mathbb{Z}$.
Simply out of curiosity, and because it might be useful later, I was wondering if this extends to any finitely-generated subgroup. If $A=\{a_{1}, a_{2}, \ldots, a_{k}\}$ for some fixed elements $a_{i}\in G$, for some fixed $k\in\mathbb{Z}^{+}$, and if we wanted to check if $\langle A\rangle$ is normal in $G$, does it suffice to show that $ga_{i}g^{-1}=a_{i}^{n}$ for every generator $a_{i}\in A$? I think the answer is yes, and here is my attempt at a proof.
Suppose for all $g\in G$, for all $1\leq i\leq k$, $ga_{i}g^{-1}=a_{i}^{n_{i}}$ for some $n_{i}\in\mathbb{Z}$. Then, $\langle A\rangle = \{a_{1}^{\alpha_{1}}a_{2}^{\alpha_{1}}\ldots a_{k}^{\alpha_{k}} \mid a_{i}\in A, \alpha_{i}\in\mathbb{Z}\}$. So, taking an arbitrary element $x$ from $\langle A\rangle$, we have that
$$gxg^{-1}=g(a_{1}^{\alpha_{1}}a_{2}^{\alpha_{1}}\ldots a_{k}^{\alpha_{k}})g^{-1} = g\left[a_{1}^{\alpha_{1}} (g^{-1}g) a_{2}^{\alpha_{1}}(g^{-1}g) \ldots(g^{-1}g) a_{k}^{\alpha_{k}}\right]g^{-1} = (ga_{1}^{\alpha_{1}}g^{-1})(g a_{2}^{\alpha_{1}}g^{-1})(g \ldots g^{-1})(g a_{k}^{\alpha_{k}}g^{-1}) $$
One of the lemmas of the homework problem was that $gx^{k}g^{-1}=(gxg^{-1})^{k}$ for $k\in\mathbb{Z}$, and the proof was pretty straightforward. Continuing, we have that
$$= (ga_{1}g^{-1})^{\alpha_{1}} (ga_{2}g^{-1})^{\alpha_{2}} \ldots(ga_{k}g^{-1})^{\alpha_{k}} = a_{1}^{n_{1}\alpha_{1}} a_{2}^{n_{2}\alpha_{2}} \ldots a_{k}^{n_{k}\alpha_{k}} \in \langle A\rangle $$
and $\langle A\rangle$ is normal in $G$.
Therefore, if conjugation of the generators results in them raised to some power, then $\langle A\rangle$ is normal, and hence, we only need to check the generators to check for the normality of $\langle A\rangle$.
This seems sensible and seems correct, I just wanted to check it with the greater public. Also, even if it is true, is it useful? Is the best/easiest way to check for the normality of a finitely-generated subgroup to check if conjugation of the generators results in them raised to a certain power, or will this never come in handy? Thanks!
 A: Your test for proving normality of $\langle A\rangle$ does suffice (where $A=\{a_1,\ldots,a_k\}$). But it is not necessary. For example, the following property also suffices to prove $\langle A\rangle$ is normal.

For all $g\in G$ and $1\leq i\leq k$, $ga_ig^{-1}=a_j^n$ for some $j\leq k$ and $n\in\mathbb{Z}$.

In other words a conjugate of a generator might be a power of some other generator. And even this condition is not necessary. You could have $ga_ig^{-1}$ be the product of two of the generators, or arbitrary powers of two generators, etc.

Side remarks on notation and writing:

*

*When you write $A=\{a_1,a_2,\ldots,a_k\mid a_i\in G,k\in\mathbb{Z}^+\}$, this is how it reads: "$A$ is the set consisting of elements $a_1,\ldots,a_k$ where $a_i$ is any element of $G$ and $k$ is any positive integer." So this sounds an awful lot like $A$ is all of $G$. What you mean to write is that $A=\{a_1,\ldots,a_k\}$ where each $a_i$ is some fixed element of $G$ and $k$ is a fixed positive integer.


*Your characterization of normality of $H=\langle x\rangle$ reads like this: "For all $g\in G$, $H$ is normal in $G$ if and only if $gxg^{-1}=x^a$." There are two problems here. First, you are missing a quantifier on $a$, which makes the statement incomplete. So you need to add "for some $a\in\mathbb{Z}$". Second, the quantifier on $G$ is in the wrong place. If a sentence of the form "for all $g\in G$, [blah]" is true, then it means I can pick any $g$ I want, plug it into [blah], and get a true statement. But if I choose $g$ to be the identity, your sentence becomes "$H$ is normal in $G$ if and only if $x=x^a$ (for some $a\in\mathbb{Z}$)." Altogether, what you mean to write is: "$H$ is normal in $G$ if and only if for all $g\in G$ there is some $a\in\mathbb{Z}$ such that $gxg^{-1}=x^a$."
