I had a question about the uniqueness of group elements.

Let the Klein Four group be defined as the group generated by the elements ${1,a,b,c}$ such that $a^2=b^2=c^2=1$ and $ab=c$, $bc=a$, $ca=b$, and $1$ is the identity element.

Let the "Klein Five" group be defined as the group generated by the elements ${1,a,b,c,d}$ such that $a^2=b^2=c^2=d^2=1$ and $ab=c$, $bc=d$, $cd=a$, $da=b$, and $1$ is the identity element.

If I manipulate the symbols of the "Klein Five" group I defined above, I can show every element is equivalent to the identity. From $ab=c=ad$ I can see $a$ is the identity, from $bc=d=ba$ I can see $b$ is the identity, and so on. This gives that $a=b=c=d=1$. In some sense, this group doesn't seem to exist. I can't make a group such that $a \neq b \neq c \neq d \neq 1$ with the constraints above.

How do I know that the same isn't true of the Klein Four group? How do I know there isn't some set of constraints that makes the group "non-existent" for unique elements in the same way as the "Klein Five" group?

Any help would be appreciated!

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    $\begingroup$ One way is to actually construct a group satisfying these relations. Then you know the group given by the presentation has to have at least as many elements as the group you constructed. $\endgroup$ – Matt Samuel Dec 2 '15 at 2:24
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    $\begingroup$ If you can write down a multiplication table, then you have the group. $\endgroup$ – Michael Burr Dec 2 '15 at 2:46
  • $\begingroup$ If you know about direct products, take $Z_2 \times Z_2$ and verify that it satisfies the properties of the Klein 4-group, and therefore it is (isomorphic to) that group. $\endgroup$ – Bungo Dec 2 '15 at 2:50

You can simply write out the full list of elements and multiplication table of the Klein Four group, and check that it really satisfies the group axioms (and that $1$, $a$, $b$, $c$, are distinct elements and in fact are all the elements of the group). If you tried to do this with the "Klein Five" group, you would run into trouble: for any multiplication table you tried to write down with $1$, $a$, $b$, $c$, and $d$ as distinct elements which satisfied your equations, it would fail to satisfy the group axioms (by the argument you give).

In general, though, it is a very hard problem to identify whether a group presentation (i.e., a list of generators and relations between them) describes only the trivial group. In fact, it is so hard that you can prove it is unsolvable in general, in the sense that there exists no algorithm that takes a finite presentation of a group and decides whether the group is trivial (this is a variant of the so-called "word problem").

  • $\begingroup$ Interesting! I'll have to take a look at that "word problem" page, although glancing over it, it at least looks beyond my current ability to understand. Based on what you've written here, does it mean that if I can assume that all elements are distinct and in a sense "not run into problems" (like the "Klein Five" does), that the group exists? $\endgroup$ – Paul Hlebowitsh Dec 2 '15 at 17:55
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    $\begingroup$ If you can completely describe a set together with a binary operation such that: (1) you can prove it is a group, (2) you can label certain elements of the set with your generators such that the relations you want are true of those elements, and (3) you can prove that the elements you've labelled are not equal to each other, then you can say those elements are distinct. The tricky part is constructing a set such that you can actually do (1) and (3) at the same time. $\endgroup$ – Eric Wofsey Dec 2 '15 at 19:10
  • $\begingroup$ One way to do this is to write down an entire multiplication table and check everything by brute force, as I suggest here. Another way is to find some set with a binary operation such that it is easy to show (1) by a conceptual argument (e.g., because the binary operation is composition of functions and the functions all have inverses), and then check that you can choose elements of that set such that (2) and (3) hold (this is the approach taken in Bungo's answer). $\endgroup$ – Eric Wofsey Dec 2 '15 at 19:13
  • $\begingroup$ In any case, the point is that it is not enough to just "play around" with the relations and "not run into problems"; you have to explicitly and completely write out a structure that you can prove is a group. $\endgroup$ – Eric Wofsey Dec 2 '15 at 19:14
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    $\begingroup$ So for the "Klein Five" group mentioned above, it should violate one of the group axioms. In this case, it has an identity, inverses, and it's closed, so it must violate associativity. It seems like it does, because a(bc) = ad != (ab)c = 1. So therefore, this group is not defined. $\endgroup$ – Paul Hlebowitsh Dec 2 '15 at 23:47

In case you don't relish the idea of writing down a multiplication table and verifying that it satisfies the group axioms (checking associativity in particular is tedious even for a $4 \times 4$ table, and impractical for anything much larger than that), another approach is to exhibit the Klein four group as a subgroup of an existing group.

For example, consider the set $GL_2(\mathbb R)$ of invertible $2\times 2$ matrices, which form a group under multiplication. Then verify that the set $$\left\{\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}, \begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix}, \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}, \begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix} \right\}$$ is closed under multiplication and inverses, and hence is a subgroup of $GL_2(\mathbb R)$. Then check that it satisfies the conditions which define the Klein 4-group.

  • $\begingroup$ Just to check my understanding, is this saying something like "If I have a representation of the group that satisfies all of the group rules, with every element distinct, I know the group doesn't reduce to the trivial group"? $\endgroup$ – Paul Hlebowitsh Dec 2 '15 at 17:52
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    $\begingroup$ @PaulHlebowitsh: Yes, that's the basic idea. In the example above, we see that the set is not trivial (it contains four distinct matrices), so if we can show that it is a group and that it satisfies the constraints for the Klein group, then it is (a particular representation of) that group. The advantage of "finding" the desired group as a subgroup of an existing group is that we don't have to prove things like associativity which are inherited from the "parent" group. $\endgroup$ – Bungo Dec 2 '15 at 18:06
  • $\begingroup$ Thank you very much! I feel as though I understand what is happening here better now. $\endgroup$ – Paul Hlebowitsh Dec 2 '15 at 18:45

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