hEorpaigh3tR

2022-11-24

Generators of a free group

If G is a free group generated by n elements, is it possible to find an isomorphism of G with a free group generated by n-1 (or any fewer number) of elements?

If G is a free group generated by n elements, is it possible to find an isomorphism of G with a free group generated by n-1 (or any fewer number) of elements?

Dillan Foley

Beginner2022-11-25Added 9 answers

Here is another approach. Let $G$ be a free group on $m$ generators, and let $H$ be a free group on $n$ generators. There are exactly ${2}^{m}$ homomorphisms from $G$ to a group of order two, since each generator can be mapped in two ways. Likewise, there are ${2}^{n}$ homomorphisms from $H$ to a group of order two.

If $G$ and $H$ are isomorphic, then they have the same number of homomorphisms to a group of order two. Therefore ${2}^{m}={2}^{n}$, which implies $m=n$

If $G$ and $H$ are isomorphic, then they have the same number of homomorphisms to a group of order two. Therefore ${2}^{m}={2}^{n}$, which implies $m=n$

Finn Mosley

Beginner2022-11-26Added 1 answers

No, if ${F}_{n}$ and ${F}_{m}$ are isomorphic, then their abelianizations are also isomorphic, which only happens when n=m (for example, reduce mod p and count points).

How to find out the mirror image of a point?

How many 3/4 Are in 1

Convert 10 meters to feet. Round your answer to the nearest tenth

6. Reduce the following matrix to reduced row echelon form:

Let v be a vector over a field F with zero vector 0 and let s,T be a substance of V .then which of the following statements are false

Describe Aut(Zp), the automorphism group of the cyclic group Zp where p is prime. In particular find the order of this group. (Hint: A generator must map to another generator)

Let a and b belong to a ring R and let m be an integer. Prove that m(ab) = (ma)b = a(mb)