Let F_n be the free group on n elements. Give (a) an injective homomorphisms fro

Question

Let F_n be the free group on n elements. Give
(a) an injective homomorphisms from F_2 to F_3;
(b) a surjective homomorphism from F_3 to F_2;
(c) an injective homomorphism from F_3 to F_2.
Are F_2 and F_3 isomorphic? Let F_n be the free group on n elements. Give
(a) an injective homomorphisms from F_2 to F_3;
(b) a surjective homomorphism from F_3 to F_2;
(c) an injective homomorphism from F_3 to F_2.
Are F_2 and F_3 isomorphic? Let F_n be the free group on n elements. Give
(a) an injective homomorphisms from F_2 to F_3;
(b) a surjective homomorphism from F_3 to F_2;
(c) an injective homomorphism from F_3 to F_2.
Are F_2 and F_3 isomorphic?

Explanation / Answer

a)Note that is surjective since inverses exist in a group, and is injective since inverses are unique.

To b), note that since : G G is a bijection,

to prove it’s an isomorphism it suffices to show it’s a homomorphism. To that end, note that if is a homomorphism then

for a, b G we have ab = (b a 1 ) 1 = (b 1 a 1 ) = (b 1 )(a 1 )

= (b 1 ) 1 (a 1 ) 1 = ba, so G is abelian.

Conversely, if G is abelian, then for all a, b G we have (ab) = (ab) 1 = b 1 a 1 = a 1 b 1 = (a)(b), so is a homomorphism.

C)Since Z is an (infinite) cyclic group with generator 1 Z, any homomorphism : Z Z is determined by a choice of image (1) Z. For n Z, let n : Z Z be the homomorphism with n(1) = n, then for any a Z, n(a) = n · a. Since multiplication of integers distributes over addition, we see that each n is in fact a homomorphism, so the collection {n : Z Z : n Z} constitutes all homomorphisms Z Z. Now we have three cases: • 0 : Z Z is the constant zero map, i.e. the trivial homomorphism.

It is neither injective nor surjective.

• n : Z Z for n 0 are all injective since a ker(n)

na = 0 a = 0, since we’re assuming n 0.

Thus ker(n) = {0},

and thus is injective. • 1 : Z Z is the identity map, which is an isomorphism and 1 : Z Z is the “minus” map, which is an isomorphism, since Z is abelian.

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