When can you replace the generators of a group with conjugates?

Question

I am under the (perhaps false) impression that it is sometimes possible to replace a generator of a (non-abelian) group with its conjugate, and have the new generators generate the same group.

For which groups is this true?

Answer 1

If $S = \{x_1,\dotsc,x_n\}$ is a generating set for a group $G$, then $\{g^{-1}x_1g,\dotsc,g^{-1}x_ng\}$ is a generating set for all $g \in G$. This latter generating set is the image of $S$ under the inner automorphism $h \mapsto g^{-1}hg$. However, one cannot in general replace the $x_i$ by conjugates by different elements of $G$.

Assuming you know about free groups, consider $G$ a free group of rank 2 generated by $a$ and $b$. Then $\{a,baba^{-1}b^{-1}\}$ is not a generating set for $G$. It turns out that if $\{x_1,\dotsc,x_n\}$ are a free basis for $F_n$ a free group of rank $n$, it's a mildly delicate question for which $g_i$ the set $\{g_1^{-1}x_1g_1,\dotsc,g_n^{-1}x_ng_n\}$ is another free basis for $F_n$. (The answer is called Whitehead's algorithm.)

Here's an obvious sufficient condition that can be practical to check by hand but difficult to use in the abstract. Suppose $S = \{x_1,\dotsc,x_n\}$ is a generating set for $G$, and that $g \in \langle x_2,\dotsc,x_n\rangle$. Then $\{g^{-1}x_1g,x_2,\dotsc,x_n\}$ is again a generating set for $G$.