# Free groups and subgroups

• Jan 12th 2013, 05:18 AM
I-Think
Free groups and subgroups
Let \$\displaystyle F_2\$ be the free group on two generators \$\displaystyle F_2=<x,y>\$. Prove that the free group on 3 generators \$\displaystyle F_3\$ is a subgroup of \$\displaystyle F_2\$.

Not completely sure how to go about doing this. So\$\displaystyle F_3=<x,y,z>\$. Can I show that z can be written as the the product of \$\displaystyle x\$'s and \$\displaystyle y\$'s and their inverses?
Or is there some other method I should try?
• Jan 12th 2013, 10:56 AM
Deveno
Re: Free groups and subgroups
well you won't actually find three letters in an alphabet that only has two.

try this instead: prove that the subgroup generated by x2,xy and y2 is isomorphic to F3.
• Jan 13th 2013, 05:45 PM
I-Think
Re: Free groups and subgroups
So considering the subgroup \$\displaystyle H=<x^2,xy,y^2>\$, and \$\displaystyle F_3=<a,b,c>\$,
I'm trying the function f, where \$\displaystyle f(x^2)=a, f(y^2)=b, f(xy)=c\$, can I guarantee this will be a homomorphism?
• Jan 13th 2013, 06:03 PM
Deveno
Re: Free groups and subgroups
it seems pretty obvious it will be, we just need to check on all 9 products of the generators, right? (once you try to do this, i think it will become clear).

the real question is: is this an ISOMORPHISM?
• Jan 14th 2013, 06:35 PM
LoidaWard
Re: Free groups and subgroups
find three letters in an alphabet that only has two.

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