Let $F$ be the free group on $\{x_i\}_{i=1}^\infty$, and let $H \leq F$ be a subgroup with $\langle H \cup \{x_1\} \rangle = F$. Must there be a free basis $B$ of $F$ for which $B \cap H \neq \emptyset$ ?

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    $\begingroup$ Here is an idea that may work: let $H=\langle x_i^2, x_1 x_i^3 x_1^{-1} \mid i \ge 2\rangle$. Then the subset $S=\{ x_i^2, x_1 x_i^3 x_1^{-1} \mid i \ge 2\}$ freely generates $H$ and $F=\langle H,x_1 \rangle$. It seems plausible that $H$ contains no primitive elements of $F$. Perhaps Whitehead's algorithm could be used to show it? $\endgroup$ Oct 18 at 13:01

1 Answer 1


Assuming $x_1 \notin H$, we can construct a basis for F as follows:

for each j,

if $x_j \notin H$, we set $b_j=x_j$ (in particular, $b_1=x_1$);

if $x_j \in H$, set $b_j=x_1x_j$.

Then, $B=\{b_1, b_2, b_3,...\}$ is a basis for $F$ not containing any element of $H$.

It seems here that the condition $\langle H\cup \{x_1\}\rangle=F$ is unnecessary. We only need $x_1\notin H$.

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    $\begingroup$ The question asks whether there always exists a basis that intersects $H$, not whether there exists a basis disjoint from $H$. $\endgroup$ Apr 23, 2018 at 7:24

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