Solve Group Homomorphisms: Show (i) & (ii) Can Hold But Not Be Unique

[moo5003]

Homework Statement

(i) Every group-theoretic relation p=q satisfied by (a,b,c) in G a group is also satisfied by (x,y,z) in F a group.

(ii) There exists a homomorphism between G and F a->x b->y c->z.

Problem:
Show by example (i) can hold and (ii) cannot.
Show (i) can hold and (ii) can hold but not be a unique homomorphism.

The Attempt at a Solution

I'm having trouble trying to solve the first part of this question. Any help would be appreciated.

As for part 2.

Let G = (ZxZxZxZ, +) e_i is the identity in all spots except ith spot of the 4-tuple, 1 otherwise. EX: e_1 = (1,0,0,0)
a = e_1
b = e_2
c = e_3

Its pretty clear that any relation p satisfied by a,b,c must be equal to the form:
A_1(a) + A_2(b) + A_3(c) where A_i is the # of times you add or subtract.

Let F = (ZxZxZxZ, +)
x = e_1
b = e_2
c = e_3
e_4 -> e_4

Obviously this is one homomorphism were (i) and (ii) both hold

x = e_1
b = e_2
c = e_4
e_4 -> e_3

This is another homomorphism were (i) and (ii) both hold. Thus the homomorphism is not unique.

 

[Hurkyl]

e_i is the identity in the ith spot of the 4-tuple.

What goes in the other three spots? And just guessing, but I don't think you meant to put zero in the i-th spot.

 

[moo5003]

Sorry, you are correct. I was thinking 1 was the identity for some reason. I changed the def of e_i so the intended element appears.

 

[moo5003]

Last Bump, I would really like to hear what people have to say about this. I'm stuck and though I get to choose which homework problems I do, I would really like to finish this one as its a main lemma about free groups is the basis for 4/5 of the chapters in the class. Any help would be appreciated. Otherwise I won't turn this in and go to office hours. I appreciate any help you can provide. (Namely part I)

 

[StatusX]

The most general relation between 3 elements is of the form:

a^{p_1} b^{q_1} c^{r_1}...a^{p_n} b^{q_n} c^{r_n} = e

where p_i,q_i,r_i are integers (possibly zero or negative) and n is a positive integer. Now it's clear that if there's a homorphism \phi:G \rightarrow F taking (a,b,c) to (x,y,z), then x,y,z must satisfy all the relations that a,b,c do (just apply \phi to the equation above). So the question is asking you about the converse of this fact.

First consider what happens when G is generated by a,b,c. If we define \phi so that it takes (a,b,c) to (x,y,z), we can restrict the value of \phi on any element in G by the property that \phi is a homorphism, eg, if g=abc, then we must have \phi(g)=\phi(a)\phi(b)\phi(c). However, for this to be well-defined, it must be the case that if g can be written in two ways in terms of a,b,c, these must map to the same element. But it's easy to see this is equivalent to the condition that any relation satisfied by (a,b,c) is also satisfied by (x,y,z). So, in this case, our definition of \phi is good, and it's easy to check that it is the unique homorphism with the desired properties.

So to find counterexamples, you must look to groups that aren't generated by (a,b,c). I'll let you work on this, but hopefully it should be clear what to look for.