Is it true that the free group $F_k$ on $k<\infty$ generators requires at least $k$ elements to generate. I.e. does every set which generates $F_k$ have cardinality at least $k$?
Asked
Active
Viewed 107 times
1 Answers
0
Yes. Sketch: any fewer than $k$ elements can't even generate the abelianization $\mathbb{Z}^k$.
Qiaochu Yuan
- 468,795
-
Thanks :) I guess I can show that $\mathbb{Z}^k$ requires at least $k$ generators by embedding into $\mathbb{Q}^k$ and using linear algebra. Is there a more elementary way? – Matt Gallagher Sep 18 '14 at 19:05
-
@Matt: you can look at the quotient $\mathbb{F}_2^k$ and use linear algebra over $\mathbb{F}_2$. – Qiaochu Yuan Sep 19 '14 at 20:41