Classify $\mathbb{Z_6} \times \mathbb{ Z_{24}} / \langle(3,2)\rangle$ according to fundamental theorem of finitely generated abelian groups.

The order of $G/H = 12$

So it can be isomorphic to $\mathbb{Z_3} \times \mathbb{Z_4}$ or $\mathbb{Z_3} \times \mathbb{Z_2} \times \mathbb{Z_2}$

$(0,1)$ has order of 4, $(1,0)$ has order of 12, $(1,1)$ has order of 24

so I'm picking group which has order 4, in this case isomorphic to $\mathbb{Z _3} \times \mathbb{Z_4}$.

a) Is this correct and any other better way to do it?

b) If I were to use $3a+2b =0$, how can I find the isomorphic group (not too sure how to do it)?

Thanks for the help folks

  • 1
    $\begingroup$ I think it's both correct and a good approach. $\endgroup$ – Greg Martin Nov 5 '14 at 8:33

The group in question is generated by $a,b$ satisfying $6a=0$, $24b=0$, $3a+2b=0$. In matrix form, this is $$ \pmatrix{ 6 & 0 \\ 0 & 24 \\ 3 & 2 } $$ Its Smith normal form is $$ \pmatrix{ 1 & 0 \\ 0 & 12 \\ 0 & 0 } $$ which proves that the group is $C_{12}$.

Following the Smith reduction (or by hand), we get that the group is generated by $a,c$ with $c=a+b$ and then $a+2c=0$, $12c=0$. Therefore, $c=a+b$ is a generator of the group.


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