Isomorphism class of quotient group

Question

I am dealing with a problem of Algebraic Topology and I have reached to this:

I have a group $G$ (abelian) and I have to find the possible isomorphism classes of $G$, if :

$G/\mathbb Z_2 \simeq \mathbb Z_2$

Now I am not sure how to proceed. Considering that $G$ is abelian and $\mathbb Z_2$ is a subgroup of $G$, from Lagrange Theorem, I find that $G$ has 4 elements, so it has to be isomorphic with $\mathbb Z_4$.

Is that true? Moreover, are there any other possible isomorphisms? Could it be , for instance, $G=\mathbb Z_2 \oplus \mathbb Z_2$ ?

Thanks in advance for your time.

Answer 1

Yes $G$ could either be cyclic of order $4$ or it could be Klein 4 $(Z_{2} \times Z_{2}$). In each case, the quotient by $Z_{2}$ is order $2$ and that must be $Z_{2}$ as this is the only group of order $2$.

So there are two possible isomorphism classes, the ones you describe.

Answer 2

There are exactly two isomorphism classes of groups of order $4$.

If $G$ has an element $x$ of order $4$, then $G$ is generated by $x$ (since $G$ has order $4$). Hence $G$ is cyclic, isomorphic to $\mathbb{Z}_4$.

Otherwise, since the order of an element of $G$ divides the order of the group, any element has order $1$ or $2$. The only element of order $1$ is the neutral element $1_G$. Pick two distinct elements $x,y\in G\setminus\{1_G\}.$

Then $xy$ is the missing element: indeed, if $xy=1_G$, then $xy^2=x=y$, contradiction. Similarly, $xy=x$ is impossible since it would imply $x=1_G$, and $xy=y$ is impossible too.

Hence $G=\{1_G,x,y,xy\}$. Notice that $(xy)^2=1_G$ by a remark above.

Now you can check that you the bijection $f:\mathbb{Z}_2\times\mathbb{Z}_2\to G$ sending $(0,0),(1,0),(0,1),(1,1)$ respectively to $1_G,x,y,xy$ is a group isomorphism (checking case by case), meaning that $G$ is isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$.