Suppose $G$ is a finite group and $N$ is a normal subgroup of $G$.

Suppose also that $X$ is a generating set for $G$, and that $Y$ generates $N$ as a normal subgroup of $G$ (i.e. $N$=$\langle \bigcup\limits_{g\in G} g^{-1}Yg \rangle$).

Is it true that the set $\{ v^{-1}uv \mid u\in Y,v\in X\}$ generates $N$? Clearly this is true if and only if for every $g\in G$ and $y\in Y$, $g^{-1}yg \in \langle \{ v^{-1}uv \mid u\in Y,v\in X\} \rangle$.

I don't think this is true, but haven't been able to find a counterexample.


No, it is not true.

Let $G=S_n$ and $N=G$. Then $X=\{(12),(23),(34),\dots,(n-1\,n)\}$ generates $G$, and $Y=\{(12)\}$ normally generators $N$ in $G$.

But $(12)$ commutes with $(12),(34),\dots,(n-1\,n)$, and $(23)(12)(23)=(13)$. So these two generator $S_3$.

If you absolutely require a proper normal subgroup, you can choose:

$$G=S_n\times \mathbb Z/2\mathbb Z,\,N=S_n\times\{0\}$$

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