Determining the structure of the abelian group, integral matrix

Question

I am revising for my upcoming university exams and I have a past exam question that I am finding particularly challenging...

a) Consider the integral matrix

$$R=\begin{bmatrix} 2 & 2 & 4 & 2\\ 4 & 4 & 8 & 5\\ 6 & 12 & 12 & 8\\ 4 & 10 & 8 & 6\\\end{bmatrix}$$

Determine the structure of the abelian group given by generators and relations

$$A_R := \langle a_1,a_2,a_3,a_4 \mid R.a=0 \rangle$$

b) Determine the number of elements of order 2 in $A_R$.

Any help with this would be very much appreciated!

Thanks in advance!

Answer 1

You know that invertible matrix operations do not change the range of the matrix, thus $$A_R = \mathrm{cok}(R \colon \mathbb Z^4 \to \mathbb Z^4)$$ remains the same. Subtract row one once from row two and four and twice from row three. You end up with the matrix $$ \begin{pmatrix} 2 & 0 & 0 & 0 \\ 4 & 0 & 0 & 1 \\ 6 & 6 & 0 & 2 \\ 4 & 6 & 0 & 2\end{pmatrix} $$ Now you make the obvious column operations to end up with

$$ \begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 6 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} $$ The result is clearly $$ A_R = \mathbb Z \oplus \mathbb Z/2 \mathbb Z \oplus \mathbb Z/6\mathbb Z $$