6
$\begingroup$

I will try and show that if our group $G$ is not cyclic then it is isomorphic to Klein's four-group $V$.

So suppose $G$ is not cyclic. Then its elements, excluding $e$, are of order $3$ or $2$. Supposing there is an element of order $3$ gives three distinct elements $g, g^2, g^3$. By hypothesis there exists $h \in G$ such that $h$ is distinct from these three elements. So we can make $gh$ which must be equal to one of the elements previously mentioned. Setting this equal to the various elements:

\begin{align} &gh=g \Rightarrow h=e && gh=g^2 \Rightarrow h=g & \\ &gh=h \Rightarrow g=e, && gh=g^3 \Rightarrow h=g^2 & \end{align} Each of which is a contradiction. Therefore all the elements are of order two and we can get an isomorphism between $G$ and $H \times H$ where $ H$ is cyclic of order two. But this direct product is isomorphic to Klein's four-group so $G$ is isomorphic to $V$.

Is this correct?

Also it is just given in the book that $V$ is isomorphic to this product; how do I show this without checking that $f(gh)=f(g)f(h)$ for all combinations?

  • 2
    Given the detail you went into on showing there can be no element of order 3 (rather than just citing 'the order of an element must divide the order of the group'), simply saying 'we can get an isomorphism between G and $\mathbb{Z}_2\times\mathbb{Z}_2$' seems to be shortcutting a lot. I'd suggest simply building the multiplication table for the group following the basic logic 'consider the elements $f, g, h$ not the identity in $G$; then $f^2=g^2=h^2=e$ since they all must be of order 2. Now, $fg=\ldots$' - once you've built it you can just compare it to the multiplication table for $V$.2012-07-01
  • 0
    There cannot be **an** element of order $3$ ($x\neq e$).2012-07-01

2 Answers 2