Question

If G is a group and $\mathbf{[}\mathbf{G}\mathbf{:}\mathbf{\text{}}\mathbf{G}\mathbf{/}\mathbf{\mathbb{Z}}\mathbf{\left(}\mathbf{G}\mathbf{\right)}\mathbf{]}\mathbf{=}\mathbf{4}$, prove that$\mathbf{\text{}}\mathbf{G}/\mathrm{\mathbb{Z}}\left(\mathrm{G}\right)\cong {\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$.

Short answer

It is proved that$\text{}\mathrm{G}/\mathrm{\mathbb{Z}}\left(\mathrm{G}\right)\cong {\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$.

Step-by-step solution

Theorem used for proof.

Theorem 8.8

Every group of order 4 is isomorphic to either${\mathrm{\mathbb{Z}}}_4$or ${\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_4$.

Proving that ​G/ℤ(G)≅ ℤ2×ℤ2

As it is given that $[\mathrm{G}:\text{}\mathrm{G}/\mathrm{\mathbb{Z}}\left(\mathrm{G}\right)]=4$, therefore the order of $\text{}\mathrm{G}/\mathrm{\mathbb{Z}}\left(\mathrm{G}\right)$is 4.

According to the theorem 8.8 ,

$\text{}\mathrm{G}/\mathrm{\mathbb{Z}}\left(\mathrm{G}\right)\cong {\mathrm{\mathbb{Z}}}_4$ or $\text{}\mathrm{G}/\mathrm{\mathbb{Z}}\left(\mathrm{G}\right)\cong {\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$

Let’s assume $\text{}\mathrm{G}/\mathrm{\mathbb{Z}}\left(\mathrm{G}\right)\cong {\mathrm{\mathbb{Z}}}_4$

As we know, if a factor group is cyclic then the overall group must be Abelian.

Since we assumed${\mathrm{\mathbb{Z}}}_4$is cyclic so G must be abelian. Which implies, $\mathrm{\mathbb{Z}}\left(\mathrm{G}\right)=\mathrm{G}$

So, the order of group can be given as

$\begin{array}{c}|\text{}G/\mathbb{Z}\left(G\right)|=\left|G\right|/\left|\mathbb{Z}\left(G\right)\right|\\ =\left|G\right|/\left|G\right|\\ =1\end{array}$

Since we already know that the order is 4. So, our assumption is wrong and $\text{}\mathrm{G}/\mathrm{\mathbb{Z}}\left(\mathrm{G}\right)$is not isomorphic to ${\mathrm{\mathbb{Z}}}_4$. Therefore, it must be isomorphic to ${\mathbb{Z}}_2\times {\mathbb{Z}}_2$.

Hence, it is proved that $\text{}\mathrm{G}/\mathrm{\mathbb{Z}}\left(\mathrm{G}\right)\cong {\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$.