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
    $\begingroup$ 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 $V$. $\endgroup$ – Steven Stadnicki Jul 1 '12 at 15:45
  • $\begingroup$ There cannot be an element of order $3$ ($x\neq e$). $\endgroup$ – mrs Jul 1 '12 at 15:56

An idea: let $\,G=\{1,a,b,c\}\,$ be a non cyclic group of order $\,4\,$, so we can assume $\,a,b\,$ are not powers of each other.

The question is: what is $\,ab\,$?? You should find pretty easy to show this must be $\,c\,$ and, with a few lines more, you prove both that $\,G\,$ is abelian and in fact the same (i.e., isomorphic to) the Viergruppe $\,V\,$.

Another idea: you seem to feel pretty comfortable using isomorphisms, Klein group and stuff: why won't you use directly Lagrange's Theorem and avoid all that long proof about the impossibility of an element of order 3?

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    $\begingroup$ Your first line... should that be "$a,b$ are not powers of each other"? $\endgroup$ – Arturo Magidin Jul 1 '12 at 18:04
  • $\begingroup$ Of course....yikes! Thank you @ArturoMagidin . Corrected $\endgroup$ – DonAntonio Jul 1 '12 at 19:43

I was going to ask if my proof of this is correct, so here it is, please tell me if it is fine.

Proof. Let G be a group of order 4 by the Lagrange theorem any element of G has order 1,2 or 4. If G has an element of order 4 then it generates so G is cyclic. Assume G has not elements of order 4, then G has 3 elements of order 2. Since the order of G is $2^2$ then G is Abelian (by the class equation). So G is Abelian and has 3 elements of order 2 and 1 of order 1 (the identity), it follows that G is isomorphic to the klein four group.

  • $\begingroup$ This is isn't wrong. But it's incomplete. You have to define a map to show the isomorphism $\endgroup$ – topologicalmagician May 18 at 11:27

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