Okay, I was working through some examples of groups and how to find homomorphisms and go from there to get the kernal of the group action.

For example, I am working on $\displaystyle D_8$. In case you want it here's the multiplication table:

$\displaystyle

\begin{array} {c||c|c|c|c|c|c|c|c|}

D_8 & e & r & r^2 & r^3 & s & sr & sr^2 & sr^3 \\

\hline \hline

e & e & r & r^2 & r^3 & s & sr & sr^2 & sr^3 \\

\hline

r & r & r^2 & r^3 & e & sr^3 & s & sr & sr^2 \\

\hline

r^2 & r^2 & r^3 & e & r & sr^2 & sr^3 & s & sr \\

\hline

r^3 & r^3 & e & r & r^2 & sr & sr^2 & sr^3 & s \\

\hline

s & s & sr & sr^2 & sr^3 & e & r & r^2 & r^3 \\

\hline

sr & sr & sr^2 & sr^3 & s & r^3 & e & r & r^2 \\

\hline

sr^2 & sr^2 & sr^3 & s & sr & r^2 & r^3 & e & r \\

\hline

sr^3 & sr^3 & s & sr & sr^2 & r & r^2 & r^3 & e \\

\hline

\end{array}

$

This is the group of symmetry operations on a square. Defining the points of the square by the coordinates (1, 0), (0, 1), (-1, 0), and (0, -1). r is the rotation of the square by 90 degrees and s is the reflection of the corners over the x-axis.

The group has 5 elements of order 2 and a cyclic subgroup of order 4. It is non-Abelian.

Now, I want to find a homomorphism $\displaystyle \varphi : D_8 \times H \rightarrow H$. (H is a set, not necessarily a group.) My problem is finding an H such that |H| is not 8 and $\displaystyle ker ( \varphi )$ is non-trivial. Just to be clear, my text gives the following definition: $\displaystyle ker( \varphi ) = \{ d \in D_8 ~ | ~ d \cdot h = h ~\forall ~ h \in H \} ~ \forall d \in D_8$, where $\displaystyle d \cdot h$ is a group action. Now, I know that if H is a subset of $\displaystyle D_8$ then $\displaystyle ker( \varphi ) = {e}$ is trivial so I have to look for a set H not contained in $\displaystyle D_8$ but I can't seem to find one. Yes, isomorphisms of $\displaystyle D_8$ exist but I'm specifically looking for $\displaystyle | H | \neq | D_8 |$ and I don't know how to construct such a set H.

I thought I had this concept down years ago. I think I'm missing something simple...

Any thoughts?

-Dan