#### AKG

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2. Let m and n be positive integers. Prove that there is a homomorphism from the free group generated by n generators, [itex]F_n[/itex], onto the free group generated by m generators, [itex]F_m[/itex], if and only if m

__<__n. Well if m = n, then the two are isomorphic, and this is obvious. If m < n, then the homomorphism that takes a word of [itex]F_n[/itex] and simply omits any occurences of [itex]x_{m+1},\, x_{m+2},\, \dots ,\, x_n[/itex] is a surjective homomorphism, where [itex]x_1,\, \dots ,\, x_n[/itex] are the generators of [itex]F_n[/itex]. I want to show that if m > n, then there is no homomorphism. This seems almost too obvious, but I'm having trouble proving it. I know that [itex]F_n[/itex] and [itex]F_n[/itex] are isomorphic if and only if m = n, and I have the isomorphism theorems. So I know that [itex]F_m[/itex] is isomorphic to [itex]F_n / \mathop{\rm Ker}(\phi )[/itex] where φ is the surjective homomorphism that was proven to exist in the first part of this proof. If there is a surjective homomorphism from [itex]F_m[/itex] onto [itex]F_n[/itex], then by the previous sentence, there is one from [itex]F_n / \mathop{\rm Ker}(\phi )[/itex] onto [itex]F_n[/itex]. Let:

[tex]\psi \, :\, F_n / \mathop{\rm Ker}(\phi ) \to F_n[/tex]

be this homomorphism. Then [itex]F_n[/itex] is isomorphic to:

[tex](F_n /\mathop{\rm Ker}(\phi ) )/\mathop{\rm Ker}(\psi )[/tex]

It seems like this should be so simple, but I can't seem to prove it.

3. Check that {x, y|xyx

^{-1}y

^{-1}} is a presentation for [itex]\mathbb{Z} \times \mathbb{Z} = \mathbb{Z}^2[/itex]. That is to say that [itex]\mathbb{Z}^2[/itex] is isomorphic to [itex]F_2/N[/itex] where N is the smallest normal group containing [itex]xyx_{-1}y^{-1}[/itex]. Well the function defined by:

[tex]\phi(x^{a_1}y^{b_1}\dots x^{a_n}y^{b_n}) = (\sum _{i = 1} ^n a_i , \sum _{i = 1} ^n b_i)[/tex]

is a surjective homomorphism from [itex]F_2[/itex] to [itex]\mathbb{Z}^2[/itex]. I need to show that it's kernel is N. I know that N is contained in the kernel, so I need to show that if g is in the kernel, then it is in N, that is, if I have any word [itex]x^{a_1}y^{b_1}\dots x^{a_n}y^{b_n}[/itex], then I can express it as some conjugate of products of conjugates of products... of conjugates of xyx

^{-1}y

^{-1}. Or maybe there's a better way?

4. Show that F

_{2}x F

_{2}is not a free group. That is to say that there is no set of generators such that there is a unique word for every non-identity element of the group. Suppose there were such a set of generators. Then (x,y) would only be the result of a unique word, but (x,y) = (x,0)(0,y) = (0,y)(x,0), so if the word for (x,0) is w, and the word for (0,y) is w', then ww' = w'w. w is not equal to w', neither of w and w' is a power of the other, and neither is the empty word. If w and w' are words of the same length, then w = w', which is impossible, so they are of different length. Define |w| as the length of the word w, and without loss of generality, assume |w| > |w'|. The last |w'| letters of w must be w', as must the first |w'|. Suppose let v = (w')

^{-1}w. Then we get:

vw' = w = w'v

Now |v| is not equal to |w'|, otherwise v = w' since v is the last |v| letters of w and w' is the last |w'| letters of w, and this would imply that w = (w')², which we already determined is impossible. Continually taking the shorter word, and finding the word that results from removing the shorter word from the front of the longer word, I think this process will eventually lead us to some sort of contradiction, namely that one of (x,0) and (0,y) is a power of the other. Is this right?

5. Suppose we have a group action on a tree [itex]\Gamma[/itex]. If the group element

*g*fixes two vertices,

*u*,

*v*of [itex]\Gamma[/itex], prove that

*g*must leave all of the geodesic [itex]\vec{uv}[/itex] fixed.

I'm not exactly sure what's being asked here. An action of G on [itex]\Gamma[/itex] is technically a pair of actions, one on the vertices of the tree, and the other on the edges of the tree. The action on the edges maps each element of g to a permutation of the edges. We can simply look at g as a bijective mapping from A to A, where A is the set of edges for the tree. But a geodesic is a path. It is made up of elements of A, but it is not itself an element of A, so treating g as a mapping whose domain is A, it doesn't technically make sense to speak of g leaving a geodesic fixed. Is there some aspect of the definition of an action on a tree that I'm overlooking, because as far as I can tell, the question doesn't even make sense.