Find the left cosets of subroups of $S_3$ So I am struggling to understand the definition of a coset.
If I have the following symmetric group $S_3=\{1, \sigma, \sigma\tau, \sigma\tau^2, \tau, \tau^2\}$, where $\sigma$=$\left(\begin{array}{ccc}1 & 2 & 3 \\ 1 & 3 & 2\end{array}\right)$ and $\tau$=$\left(\begin{array}{ccc}1 & 2 & 3 \\ 2 & 3 & 1 \end{array}\right)$.
I am given the subgroups $H=${$1,\sigma$} and $K=${$1,\tau,\tau^2$}
How would I find all the left cosets of these subgroups? Thanks
 A: We note that $S_3 = \{e, (1 \, 2), (1 \,3), (2 \, 3), (1 \, 2 \,3), (1 \,3 \,2)\}$. We're given that $H = \{e, (2 \,3)\}$, and $\tau = \{e, (1 \,2 \,3), (1 \,3 \,2)\}$
For a group $G$ with $H \leq G$, the definition of a left coset of $H$ in $G$ is given by $$gH = \{gh \,|\, h \in H\}.$$ By Lagrange's Theorem, we know there should be $[S_3 : H] = {|S_3| \over |H|} = {3! \over 2} = 3$ distinct left cosets of $H$ in $G$. We only need to find three distinct cosets and then we're done. Well, for $e \in S_3$, $$eH = \{eh \,|\, h\in H\} = \{h \,|\,h \in H\} = H.$$ So we found one left coset, namely $H$ itself. Now we need two more. For $(1\,2) \in S_3$, we have $$(1 \, 2)H = \{(1 \, 2)e, (1 \,2) (2\,3)\} = \{(1\,2),(1 2 3)\}.$$
This is distinct because we haven't seen it before. So $(1\,2)H$ is another distinct left coset. Now try for $(2 \,3) \in S_3$. $$(2 \, 3)H = \{(2 \,3)e, (2 \,3)(2 \,3)\} = \{(2 \,3), e\} = H.$$
Uh-oh. This isn't distinct because we've already found this left coset. We need to try another one. Let's try $(1 \,3 \,2) \in S_3$. $$(1\,3\,2)H = \{(1\,3\,2)e,(1\,3\,2)(2\,3)\} = \{(1\,3\,2), (1\,3)\}.$$ We haven't seen this left coset before. Thus, $(1\,3\,2)H$ is our final distinct left coset. And again, by Lagrange's Theorem, we know we're done. So the left cosets of $H$ in $S_3$ are $H$, $(1\,2)H$, and $(2\,3)H$.
You repeat the exact same process to find the left cosets of $\tau$ in $S_3$. In this case, by Lagrange's Theorem, the number of distinct left cosets of $\tau$ in $S_3$ is $[S_3 : \tau] = {|S_3| \over |\tau|} = {3! \over 3} = 2$. We know that for $e\in S_3$, $e\tau = \tau$. So a distinct left coset of $\tau$ in $S_3$ is $\tau$ itself. Now we just need to find one more distinct left coset. Try $$(1\,3)\tau = \{(1\,3)e,(1\,3)(1\,2\,3),(1\,3)(1\,3\,2)\} = \{(1\,3),(1\,2),(2\,3)\}.$$ Awesome. We haven't seen this left coset before. And again by Lagrange's Theorem, we're done. So the distinct left cosets of $\tau$ in $S_3$ are $\tau$ and $(1\,3)\tau$.
