I'm trying to prove that given a group $G$ and a subgroup $H$ where $H\leq G$ , and given that $aH = bH$ , then we need to check if also $Ha = Hb$ .
I'm trying to show that this is wrong with a counter example :
$G=S_{3}$ , $H={(1),(1 2)}$ , $a=(1 3) , b=(1 2 3)$
And now we check : $aH = (1 3)H=(1 3) ( (1),(1 2))={(1 3) ,(1 3)(1 2)} = ? $
Here , for the multiplication $(1 3)(1)$ : 1 goes to 1 and 1 goes to 3 , then the result is 1 goes to 3 .
For the second multiplication $(1 3)(1 2)$ : 1 goes to 2 , and then 1 goes to 3 ? how do I calculate this one ?
The same for bH :
$bH=(1 2 3)H=(1 2 3) ( (1),(1 2))={(1 2 3) ,(1 2 3)(1 2)} = ? $
How do I calculate the multiplication of $(123)(12)$ ?
Regards
I believe the precise term you're looking for is composition of permutations. It works exactly like composition of functions: we look at each element's image in the first permutation, then carry that image through to the next permutation, and see where it is sent there.
For the second example, $1$ would go to $2$, and then we look at $2$'s image in the next permutation, which is $3$. This is assuming you're composing right to left, which is how your examples have been going.
Now $2$ goes to $1$ in the right permutation, and $1$ goes back to $2$ in the left permutation, so $2$'s image in the composition of both permutations is $2$. Finally, $3$ is fixed by the first permutation, so we look at where $3$ is sent in the second permutation - it is mapped to $1$. Now, we write our result as one permutation:
$(123)(12) = (13)(2)$, or just $(13)$.
For the first, we similarly calculate $(13)(12)$. Compactly writing the process: $1 \rightarrow 2 \rightarrow 2, 2 \rightarrow 1 \rightarrow 3, 3 \rightarrow 3 \rightarrow 1$, so we have that $(13)(12) = (123)$.