Uniqueness of elements in Klein Four and "Klein Five" group 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!
 A: 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").  
A: 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.
