7
$\begingroup$

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 $B \cap H \neq \emptyset$ ?

$\endgroup$
1
$\begingroup$

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 $x_1\notin H$.

$\endgroup$
1
  • 1
    $\begingroup$ The question asks whether there always exists a basis that intersects $H$, not whether there exists a basis disjoint from $H$. $\endgroup$ Apr 23 '18 at 7:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.