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(a) Given the general linear group (over the integers) GL(3,Z), show the subgroup H generated by $\begin{bmatrix}0 & 1 & 0\\ -1 & -1 & 0\\ 0 & 0 & 1\end{bmatrix}$ and $\begin{bmatrix}0 & 1 & 1\\ -1 & -1 & 0\\ 0 & 0 & 1\end{bmatrix}$ is infinite.

(b) Given the presented group $ G_3=<x,y|x^3, y^3, (xy)^3>$ , construct a surjective homomorphism from G onto H. Thus it shows $G_3$ is infinite.

My approach: For the firsth part of the problem I aim to construct a sequence of infinite matrices in H, but all sequences I tried terminated to be the identity matrix.

For the second part it's clear that the given generators of H satisfy the relations of G, so by Von Dyck's theorem H is aquotient of $ G_3$, but does this guarentee the existence of such surjective homomorpshism? In general, if given an abstractly presented group with relations which are all satisfied by generators of another known group, is there always a surjective homomorphism between the two groups which preserves the corresponding generators?

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up vote 5 down vote accepted

The trick for the first part is that the matrices lie in $\mathbb{Z}^2 \rtimes \text{GL}(2,\mathbb{Z})$. So we are done if we can construct an element with trivial part in $\text{GL}(2,\mathbb{Z})$ and non-trivial part in $\mathbb{Z}^2$.

Now, think of your matrices as elements of an affine group. The upper left $2 \times 2$ submatrix is the linear part. Note that it is the same in both matrices. Thus, compute $ab^{-1}$ (or $a^{-1}b$) – where $a$ is the first and $b$ the second matrix – to get an element with the desired properties.

For the second part: Yes, this always works. Whenever you have a finitely presented group $G = \langle x_1,\ldots,x_n | r_1(x_1,\ldots,x_n),\ldots,r_m(x_1,\ldots,x_n)\rangle$ and a group $H = \langle h_1,\ldots,h_n \rangle$ with $r_j(h_1,\ldots,h_n) = 1_H$ for all $1 \le j \le m$, $x_i \mapsto h_i$ extends to a homomorphism (which is obviously surjective). This follows from the universal property of free groups: Let $G = F(X)/R$ and $f: X \rightarrow H : x_i \mapsto h_i$. Then there exists a homomorphism $\varphi: F(X) \rightarrow H$ and, because $r_i(h_1,\ldots,h_n)=1_H$, we get $R \subset \text{ker}\varphi$. Finally, use the fundamental theorem on homomorphisms.

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