In response to the below problem:
How many homomorphisms from $K_4\rightarrow D_3.$
Have doubt:
Am not clear, why is a map of order $2$ element $\in K_4,$ possible to $e?$
The group table for $K_4$ is:
\begin{array}{|c|c|c|} \hline
&e&x&y&xy\\ \hline
e&e&x&y&xy \\ \hline
x&x&e&xy&y \\ \hline
y&y&xy&e&x \\ \hline
xy&xy&y&x&e \\ \hline
\end{array}
In $K_4,$ the order of all non-trivial elements is $2,$ and product of any two such elements is another non-trivial element in the group.
Also, any two elements commute with each other.
Say, $xy = yx.$
So, must map $\phi(a),$ for non-trivial element $a\in K_4,$ to an order $2$ element in $D_3.$
In $D_3,$ all reflection elements are of order $2.$
The two non-trivial rotation elements have order $3,$ as:
$\langle R\rangle
=\{R, R^2, e\},$
$\langle R^2\rangle
=\{R^2, R, e\}.$
The group table for $D_3$ is: \begin{array}{|c|c|c|c|c|c|} \hline &e&R&R^2&T&TR&TR^2\\ \hline e&e&R&R^2&T&TR&TR^2 \\ \hline R&R&R^2&e&TR&TR^2&T \\ \hline R^2&R^2&e&R&TR^2&T&TR \\ \hline T&T&TR&TR^2&e&R&R^2 \\ \hline TR&TR&TR^2&T&R^2&e&R \\ \hline TR^2&TR^2&T&TR&R&R^2&e \\ \hline \end{array}
So, can map any non-trivial element $\in K_4,$ to any reflection element $\in D_3.$
But, need $\phi(a)\phi(b)= \phi(b)\phi(a),$ for $a,b\in K_4.$
But, that is impossible in $D_3,$ if $\phi(a)\ne \phi(b),$ unless $\phi(a)=\phi(b),$ or $\phi(b)= e.$
Say, $TR\cdot TR^2= R,$ but $TR^2\cdot TR = R^2.$
$TR\cdot T= R^2,$ but $T\cdot TR = R.$
Hence, have $3$ choices for any element $a$ to map to.
Corresponding to each such choice, have $2$ choices for $\phi(b).$
This leads to $3\cdot 2= 6$ choices.
Next, consider the trivial map $\phi(a)=e.$ And, $\phi(b)$ can map to either $e,$ or any reflection element.
$\phi(a)= e,$ for any of the four choices available for the element $a,$
The possible choices are:
- $\phi(a=x)= e, \phi(b=y)= e,$
- $\phi(a=x)= e, \phi(b=y)= T,$
- $\phi(a=x)=e, \phi(b=y)=TR,$
- $\phi(a=x)= e, \phi(b=y)= TR^2.$
Due to the commutativity of $ab\in K_4,$ need only consider the map of one element $a,$ i.e. $\phi(a)= e. $
Else, had need to consider four more cases.
In total have $6$ non-trivial maps of generator element $\phi(a),$ and $4$ trivial maps of $\phi(a)$ in $K_4.$