If G is a group and [G:Z(G)]=4, prove that G/Z(G) isomorphic to Z2+Z2. Should I look at the cosets formed by G/Z(G) or what? I hve no clue, need help(Crying)

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- Apr 17th 2010, 03:59 PMwutanggroup theroy question
If G is a group and [G:Z(G)]=4, prove that G/Z(G) isomorphic to Z2+Z2. Should I look at the cosets formed by G/Z(G) or what? I hve no clue, need help(Crying)

- Apr 17th 2010, 04:25 PMDrexel28
You have only two choices. Since $\displaystyle [G:\mathcal{Z}(G)]=4$ we either have that $\displaystyle G/\mathcal{Z}(G)\cong \mathbb{Z}_4$ or $\displaystyle G/\mathcal{Z}(G)\cong \mathbb{Z}_2\oplus\mathbb{Z}_2$ since those are, up to isomorphism, the only groups of order four. But, $\displaystyle G/\mathcal{Z}(G)\ncong\mathbb{Z}_4$ for if $\displaystyle G/\mathcal{Z}(G)$ is cyclic then $\displaystyle G$ is abelian and so $\displaystyle \left|G/\mathcal{Z}(G)\right|=1$. It follows that $\displaystyle G/\mathcal{Z}(G)\cong \mathbb{Z}_2\oplus\mathbb{Z}_2\cong\mathcal{K}_4$

P.S. If you aren't sure why those are the only two groups you need merely note that any group of order $\displaystyle p^2$ where $\displaystyle p$ is a prime is abelian and the rest follows from the fundamental theorem of finitely generated abelian groups.