TITLE: Uniqueness of elements in Klein Four and "Klein Five" group
QUESTION [7 upvotes]: 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!

REPLY [5 votes]: 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.