Find an example, in the group $S_3$ of permutations of $\{1,2,3\}$, of elements $x,y\in S_3$ for which $x^2 = e = y^2$ but for which $(xy)^4$ $\not=$ e.
$\endgroup$ 15 Answers
$\begingroup$First of all, a quick correction: The symmetric group $S_3$ is a group of order $3! = 6$: the group of all permutations of the elements in the set $S = \{1, 2, 3\}$.
Recall that these elements are the permutations, written in cycle form here, consisting of $$S_3 = \{(1) = e, (1\; 2), (1\; 3), (2\; 3), (1\; 2\; 3), (1 \;3 \;2)\}.$$
There are three elements (permutations) in $S_3$ which have order $2$; and what this means is that, for $x\in S_3$, and $x \neq e$, but $x^2 = e$, then $x$ has order $2$.
These elements (permutations) in $S_3$ that of order two are
$$\;(1 \;2), (1\; 3), (2\; 3).\tag{ Order 2}$$ $$ \text{So}\;(1 \;2)(1 \;2) = e, \text{ and likewise, for each}$$ Any two elements of these three elements $x, y, \;\;x \neq y,\;$ are such that $x^2 = e, \; y^2 = e$, but $(xy)^2 \neq e$ and $(xy)^4 \neq e.\quad$
(Recall that $S_3$ is not abelian, so $(xy)^2 \neq x^2y^2 = e$ and $(xy)^4 \neq x^4y^4 = e.$)
Try for yourself: take any two different permutations of order two, call them $x, y$ respectively, and compose them: find $xy$; then find $(xy)^2$ and $((xy)^2)^2 = (xy)^4$, and you'll have what you need.
$\endgroup$ 1 $\begingroup$However, you have to solve the problem theoretically; I 'd like to do it for you by using GAP. Indeed, after solving such this computational problem by using theorems and facts, it is a fun doing that by machine. Here you are:
gap> S3:=SymmetricGroup(3);;
gap> o2:=Filtered(S3,x->Order(x)=2);;
gap> eo2:=Elements(o2); [(2,3), (1,2), (1,3)]
gap> for i in [1..3] do for j in [1..3] do if (eo2[i]*eo2[j])^4 <> () then Print(eo2[i]," ",eo2[j],"\n"); fi; od; od; (2,3) (1,2) (2,3) (1,3) (1,2) (2,3) (1,2) (1,3) (1,3) (2,3) (1,3) (1,2)The trick is to have $(xy)^3=e$, so that $(xy)^4=(xy)^3(xy)=e(xy)=xy\neq e$.
$\endgroup$ 2 $\begingroup$Take $x=(12)$ and $y=(23)$ ...
$\endgroup$ $\begingroup$This group is isomorphic to the 6 element dihedral group. Any element including a reflection will have order two. The product of two different such elements will be a pure nonzero rotation, necessarily of order 3.
$\endgroup$
