# Cosets on S3

• Apr 1st 2011, 01:36 PM
topsquark
Cosets on S3
First the problem:
Quote:

Let H be the cyclic subgroup (of order 2) of $S_3$ generated by $\binom {1~2~3}{2~1~3}$. Then no left coset of H (except H itself) is also a right coset. There exist an element a in $S_3$ such that $aH \cap Ha = \{ a \}$.
If I am correct the problem is simply asking about a specific example, one that does not necessarily generalize. I can easily show that, for a in H, aH = Ha = H for any subgroup, due to closure of H. And, as it happens I have showed (by construction) that for any element a of $S_3$ that is not in H, $aH \neq Ha$. (I note that this property is specific to the subgroup H. If I take a new cyclic subgroup K = <(123)> of $S_3$ that aK = Ka for all a in $S_3$.) (Note: please forgive me if my cyclic notation is off. I rarely use it.)

The last part is bugging me. $aH \cap Ha = \{ a \}$ is a true statement for all a in $S_3$! So why are they commenting on it??

-Dan
• Apr 1st 2011, 02:22 PM
Deveno
because if for every g in S3, gH = Hg, then gH∩Hg = Hg (=gH).

the fact that aH ∩ Ha = {a} ≠ aH indicates a certain defect in H.

your last statement isn't true. if h is in H, then hH = Hh = H, so hH ∩ Hh = H ∩ H = H, NOT {h}.

in particular, eH ∩ He = {e, (1 2)} (i am just writing the generator of H in cycle form), which is not {e},

and (1 2)H ∩ H(1 2) = {(1 2), (1 2)^2} ∩ {(1 2), (1 2)^2} = {(1 2), e} ∩ {(1 2),e} = {(1,2), e} which is not {(1 2)}.