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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

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    Ron: *Please*: dollar signs, not quote marks.2012-03-14
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    It depends on your conventions; Herstein multiplies them left to right; most books multiply them right to left. So $(13)(12)$ would be "first $(12)$, *then* $(13)$". This would be $1\to 2\to 2$, so $1\to 2$; $2\to 1\to 3$, so $2\to 3$; and $3\to 3\to 1$, so $3\to 1$. So $(13)(12)=(123)$.2012-03-14
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    Why [tag:abelian-groups]?2012-03-14
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    @ArturoMagidin: Thanks , but why , 3 goes to 3 (here 3→3→1) on the first hand (and not 3 goes to 1) , and on the second hand 2 goes to 1 (here 2→1→3) and not 2 goes to 2 ?2012-03-14
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    @ron: $(12)$ sends $3$ to itself, and $(13)$ sends $3$ to $1$. So the result of applying $(13)(12)$ to $3$ is: the first permutation (namely, $(12)$) sends $3$ to $3$; then you plug in this output, $3$, into the second permutation (namely, $(13)$) which sends $3$ to $1$. So, in total, $(13)(12)$ sent $3$ to $1$. $(12)$ sends $2$ to $1$; and $(13)$ sends $1$ to $3$. So the result of applying $(13)(12)$ to $2$ is: the first permutations, $(12)$, sends $2$ to $1$; then you plug in the output $1$ into $(13)$, you get $3$. So in summary, $(13)(12)$ sends $2$ to $3$ (via $1$).2012-03-14
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    Got it ! you're indeed the king! Those permutations are pretty hard to understand sometimes! much thanks ! :)2012-03-14
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    @ArturoMagidin: By the way , then (1 2 3) also means probably that I can say (2 3 1) or (3 1 2) , right ? since all the elements still are still going to the same destination !?2012-03-14
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    @ron: Yes: in cycle notation, $(a_1\cdots a_n) = (a_2\cdots a_na_1)=(a_3\cdots a_na_1a_2)=\cdots = (a_na_1\cdots a_{n-1})$.2012-03-14

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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 $(13)(12) = (123)$.

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    Thanks . I need to understand something (1 2) means that 1 goes to 2 and 2 goes to 2 ? how 1 goes to 2 and 2 goes to 1 ?2012-03-14
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    $(12)$ means $1$ goes to $2$ and $2$ goes to $1$. This is **cycle notation**, not abbreviated 2-line notation. The cycle $(a_1a_2\cdots a_n)$ represents the permutation $a_1\mapsto a_2$, $a_2\mapsto a_3$, $a_3\mapsto a_3,\ldots,a_{n_1}\mapsto a_n$, and $a_n\mapsto a_1$. So $(12)$ means "$1$ goes to $2$, $2$ goes to $1$"; $(123)$ means "$1$ goes to $2$, $2$ goes to $3$, and $3$ goes to $1$". It does **not** mean $$\left(\begin{array}{ccc}1&2&3\\1&2&3\end{array}\right)\leftarrow\text{not this}.$$2012-03-14