Is the swapping paintings of walls form a group?

Question

a . There are number of 6 ways

b. Now consider these actions $f_1$ = swapping of L and C

$f_2$ = swapping of C and R

Now we have the set as {$e,f_1, f_2, f_1f_2, f_2f_1,f_1f_2f_1$}.

Now we have actions defined, every action is reversible also and every action is deterministic and also any consecutive sequences of actions is also action because any sequence of actions is also some swap arrangement (one of the 6 arrangements)

4. Now when we apply new action we get two empty walls, one is window wall and other is right wall. So this does not form group

IS THIS CORRECT ?

Answer 1

Let the tuple $(x_1,x_2,x_3)$ denote the configuration where $x_1$ is the art on L, $x_2$ on C and $x_3$ on R. Let $f_1(x_1,x_2,x_3)=(x_2,x_1,x_3)$ and $f_2(x_1,x_2,x_3)=(x_1,x_3,x_2)$. Let the initial configuration be $(x_1,x_2,x_3)=e$. It is easy to note that $f_1\circ f_1=e$ and $f_2\circ f_2=e$. The six possible configurations are,

1. $(x_1,x_2,x_3)=e$
2. $(x_2,x_1,x_3)=f_1\circ e$
3. $(x_1,x_3,x_2)=f_2\circ e$
4. $(x_2,x_3,x_1)=f_2\circ f_1\circ e$
5. $(x_3,x_1,x_2)=f_1\circ f_2\circ e$
6. $(x_3,x_2,x_1)=f_1\circ f_2\circ f_1\circ e=f_2\circ f_1\circ f_2\circ e$

It also follows that $(f_1\circ f_2)\circ(f_1\circ f_2)=e$, $(f_2\circ f_1)\circ(f_2\circ f_1)=e$,$(f_1\circ f_2\circ f_1)\circ (f_2)=e$ and $(f_2\circ f_1\circ f_2)\circ(f_1)=e$. Closure is easy to see. Associativity follows naturally and they form a group.

Lets call the new action $f_3$. So $f_3(x_1,x_2,x_3)=(x_1,(x_2,x_3),\phi)$. Now there exists no action with which you would be able to separate $(x_2,x_3)$ pair. Say $f_2\circ f_3 \circ e=(x_1,\phi,(x_2,x_3))$. Thus you would never be able to get back to the original configuration. So the new action doesn't form a group as it doesn't have an inverse for $f_3$.