Let $H$ be the subgroup of $\Bbb Z^3$ generated by elements $(5,−4, 3), (7, 2, 3)$ and $(21, 8, 9)$. Classify the factor group $\Bbb Z^{3}/H$.

I am studying for an abstract algebra exam and am working through practice problems provided by the professor.

The problem statement is:

Let $H$ be the subgroup of $\Bbb Z^{3}$ generated by elements $(5,−4, 3), (7, 2, 3)$ and $(21, 8, 9)$. Classify the factor group $\Bbb Z^{3}/H$.

My general thought process is:

1. More generally represent the subgroup $H$.

2. Determine what the cosets look like

3. Determine what group the group of cosets is isomorphic to.

4. I noted that any element in $H$ can be written in the more general form $((5a+7b+21c), (-4a+2b+8c), (3a+3b+9c)) $ where $a,b,c \in\Bbb Z$.

5. I started to list out possible values for the first term to try and get an idea of what the cosets looked like. This quickly got out of hand as I realized just how many terms are in $H$. For example not only are all multiples of $5,7,21$ in $H$, Since $5+7 = 12$ all multiples of $12$ are in $H$. All multiples of $7-5=2$ are in $H$ etc.....

6. Clearly I can't start this part until I've figured out the first two.

My questions:

1. Have I understood the group $H$ correctly? As I type this I realize the problem statement never states what type of group this is. Was I incorrect to assume it was an additive group instead of a multiplicative group?

2. Assuming I am correct in assuming an additive group. How can I determine what my cosets look like?

Solution 1:

The subgroup $H$ is generated by $b_1=(5,−4,3),b_2=(7,2,3)$ and $b_3=(21,8,9)$. So it is generated by $b_1, b_4=b_2-b_1=(2,6,0), b_5=b_3-3b_2=(0,2,0)$. So it is generated by $b_6=b_1+2b_5=(5,0,3), b_7=b_4-3b_5=(2,0,0), b_5$. So it is generated by $b_8=b_6-2b_7=(1,0,3), b_7,b_5$. So it is generated by $b_8, b_9=b_7-2b_8=(0,0,-6), b_5$. Choose a new basis of $\Bbb Z^3$, $\{f_1=b_8, f_2=e_2, f_3=e_3\}$ where as usual $e_2=(0,1,0), e_3=(0,0,1)$. In the basis $\{f_1,f_2,f_3\}$, the last generating set of $H$ is $\{f_1,2f_2, -6f_3\}$. Therefore the group $\Bbb Z^3/H$ is the direct product of the cyclic group of order $2$ and the cyclic group of order $6$.