I'm new to group theory and struggling with understanding what is required in an example problem I'm working through.

I need to classify the following using the classification og finite abelian groups:

$$ \Bbb Z_2 \times \Bbb Z_4 / \langle (0,1) \rangle $$ $$ \Bbb Z_2 \times \Bbb Z_4 / \langle (1,2) \rangle $$ $$ \Bbb Z_6 \times \Bbb Z_8 / \langle (1,2) \rangle $$

How would I go about classifying these? Any help is greatly appreciated.


Recall that a quotient of an abelian group is again abelian. Notice that the subgroup generated by $\left\langle (0,1)\right\rangle$ in $\mathbb{Z}_2\times \mathbb{Z}_4$ has $4$ elements, thus $\mathbb{Z}_2\times \mathbb{Z}_4/\left\langle (0,1)\right\rangle$ has $2$ elements. By the classification of finite abelian groups we must have that this group is isomorphic to $\mathbb{Z}_2$.

You can proceed in this way by doing the same for the other groups. If necessary you can look at order of elements to exclude certain possibilities.

Alternatively you can use the first isomorphism theorem:

Consider the map $f:\mathbb{Z}_2\times \mathbb{Z}_4\rightarrow \mathbb{Z}_2:(\bar{a},\bar{b})\rightarrow \bar{a}$. Clearly $f$ is a surjective morphism and $\ker(f)=\left\langle (0,1)\right\rangle$, by the first isomorphism theorem we have that $$\mathbb{Z}_2\times \mathbb{Z}_4/\left\langle (0,1)\right\rangle\cong \mathbb{Z}_2.$$

If you see the proper morphism you can get the desired isomorphism immediately, but of course this method only works if you already know which group it should be.

  • $\begingroup$ How did you find that the subgroup generated by $ \langle (0,1) \rangle $ in $ \Bbb Z_2 \times \Bbb Z_4 $ has 4 elements? $\endgroup$ – jdminer Mar 13 '17 at 7:25
  • 1
    $\begingroup$ Recall that $\left\langle a\right\rangle:=\left\{a^i\mid i \in \mathbb{Z}\right\}$. Here $(0,1)^i:=(0,1)+(0,1)+\dots +(0,1)$ ($i$-times if $i\geq 0$). Thus $(0,1)^i=(0,i)$ and there are only $4$ such elements. $\endgroup$ – Mathematician 42 Mar 13 '17 at 9:19

$$\mathbb{Z}_2 \times \mathbb{Z}_4 / \langle (0,1) \rangle \cong \mathbb{Z}^2 / \langle (0,1), (2,0), (0,4) \rangle $$

You can put the generators on the right as rows in a matrix:

$$\begin{pmatrix} 0 & 1 \\ 2 & 0 \\ 0 & 4 \end{pmatrix} $$

so that the subgroup they generate is precisely the (integer) rowspace.

You can simplify this description of the subgroup by doing invertible integer row operations. e.g. you can add an integer multiple of one row to another, or swap rows, or multiply a row by $\pm 1$.

Also, you can do column operations as well; these correspond to a change of basis — i.e. isomorphisms $\mathbb{Z}^2 \to \mathbb{Z}^2$.

Your ultimate goal in doing linear algebra is to get a diagonal matrix.


The first question here is what is the operation while considering $Z_2 \times Z_4$. So, we have two options.

One is addition, or $(a,b)+(c,d)= (a+c(mod2), b+d(mod4))$, and then $<(0,1)> = \{(0,1)+(a,b)| a\in Z_2, b \in Z_4\}$ and it is a good exercise that you see by yourself that this is whole $Z_2 \times Z_4$, hence the quotient is trivial, which is not the point, I assume.

Other is multiplication, or $(a,b) \times (c,d) = (ac(mod2),bd(mod4))$, and then $<(0,1)> = \{(0,1) \times (a,b)| a\in Z_2,b \in Z_4\}=\{(0,b)|b \in Z_4\}$ which is obviously isomorphic to $Z_4$, so the quotient is obviously $Z_2$.

In other cases the thinking is similar.


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