Computing product of two cosets

Question

Let in $S_3$, $H=\{(1),(12)\}$ show that $(13)H(23)H \neq (13)(23)H$

The solution claims $(13)H(23)H=\{(132),(23),(12)\}$ however I calculated $(13)H(23)H=\{(132),(23),(12),(1)\}$

The solution seems to omit the identity from the set. Am I correct here or is the solution correct?

Answer 1

If we just take it one element at a time, we have $$ (23)H = \{(23)(1), (23)(12)\} = \{(23), (132)\}\\ H(23)H = \{(1)(23), (1)(132), (12)(23), (12)(132)\} = \{(23), (132), (123), (13)\}\\ (13)H(23)H = \{(13)(23), (13)(132), (13)(123), (13)(13)\} = \boxed{\{(132), (23), (12), (1)\}} $$ There is no reason to omit the identity element from this, and doing so would give a different subset of $S_3$. So I agree with you in this case.

Answer 2

A faster way of solving this problem than explicitly calculating cosets:

Note since $() \in H$, $(13)(23) H \subseteq (13) H (23) H$. You can see this by replacing the first $H$ on the right-hand side by $()$. So it suffices to show that the $\subseteq$ is actually a $\subsetneq$.

Since $(13)(23) H$ is a coset of $H$, and $|H| = 2$, we have $|(13)(23) H| = 2$. Moreover, the two elements in $(13)(23) H$ must be $(13)(23) = (123)$ and $(13)(23)(12) = (23)$.

So then all we need to do is think about what else could possibly be in $(13)H (23) H$. Note that $(13) H (23) H$ contains precisely the elements of the form $g := (13)h_1 (23) h_2$, where $h_1, h_2 \in H$. If $h_1 = ()$, then $g \in (13)(23) H$. So we look at elements $(13)(12)(23) h_2$. It suffices to pick one $h_2$ for which $g \notin (13)(23) H$, but you can calculate both to see what $(13)H (23) H$ actually looks like.

(And yeah, when $h_2 = (12)$, I also get $(13)(12)(23)(12) = ()$, so to answer your actual question, you are correct.)

I suppose the insight here is to try to save yourself grunt work whenever possible. Here the clever steps were noticing $|(13)(23) H| = |H|$ and $(13)(23) H \subseteq (13) H (23) H$.