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I need someone to show me how to look at a permutation and do it properly.
Do not pay attention to the right coset, I understand it. Just focus on the permutations please.
Here is my example given to me:

$$ H(12)=\{(1),(123),(132)\}(12) =\{(1)(12),(123)(12),(132)(12)\} =\{(12),(13),(23)\}$$

The $\{(1),(123),(132)\}$ is a subgroup and the $(12)$ is a group, please dont focus on that. I understand that. The way I see the end result for the $\{(1),(123),(132)\}(12)$ is that the $(1)(12)$ is $(12)$ because its being multiply by its identity. The way I see $(123)(12)$ is $(13)$ is that the $2$ cancels but I dont know the proper way to understand it. Same for $(132)(12)$ is $(23)$, I dont fully understand how to see it and determine it. Can somebody show me like saying $1$ goes to $2$, etc.,

4 Answers 4

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Based on the directions given to me on this post, I have:
$H(14)(23)$={$(1),(123),(132)$}$(14)(23)$
$\qquad$ ={$(1)(14)(23)$,$(123)(14)(23)$,$(132)(14)(23)$}
$\qquad$ ={$(423)$,$(234)$,$(324)$}
$H(13)(24)$={$(1),(123),(132)$}$(13)(24)$
$\qquad$ ={$(1)(13)(24)$,$(123)(13)(24)$,$(132)(13)(24)$}
$\qquad$ ={$(324)$,$(234)$,$(134)$
Is these right?

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Here's an example. To compute $(123)(12)$, start on the rightmost permutation with $1$. That sends $1\to2$ and then $2\to 3$. So we write $(13$ in the resulting cycle (not knowing yet that it's closed). Then we see $3 \to 3$ and then $3 \to 1$ so we've now completed the cycle. You can then check that $2\to 1$ and then $1\to 2$, so the final cycle is $(13)(2) = (13).$

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    So {$(1),(123),(132)$}$(34)$ is $(1)(34)=(34)$ and $(123)(34)=(142)$ and $(132)(34)=(124)$? Am I right? – 2017-02-14
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    Nope, both incorrect. When you're trying to compute (123)(34), think about where 1 is sent. Starting from the right, 1 goes to 1 and then 1 goes to 2. Then for two. 2 goes to 2 and thebn 2 goes to 3. Then, 3 goes to 4 and 4 goes to 4. Finally, 4 goes to 3 and 3 goes to 1. So it should be (1234). – 2017-02-14
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    I am not seeing it clearly. I thought the 3 cancels out for $(123)(34)$. Based on your last sentence, It seems like you are saying $(431)$. If you are 100% right then, i will use your answer for the $(123)(34)$ – 2017-02-14
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    What about my $(132)(34)$? – 2017-02-14
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$1,2,3$ goes to $2,3,1$ by $(123)$, and then $(12)$ permutes the first and second place, ie., $2,3,1$ goes to $3,2,1$. Altogether $1,2,3$ goes to $3,2,1$, hence $2$ is fixed and $1$ and $3$ are permuted -which is just the definition of $(13)$. So $(123)(12)=(13)$.

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    So {$(1),(123),(132)$}$(12)(34)$ is $(1)(12)(34)=(234)$ and $(123)(12)(34)=(134)$ and $(132)(12)(34)=(214)$. Am I right? – 2017-02-14
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Picking through your comments, lets work on some examples.

So $(1)(12)(34)=(234)$ and $(123)(12)(34)=(134)$ and $(132)(12)(34)=(214)$ Am I right?

No

first $(1)$ is the identity permutation. I would actually preffer to use $()$ to $(1)$ but that is more a matter of aesthetics.

$()(12)(34)=(12)(34)$

Next one.

$(123)(12)(34)$

Work right to left. Think of each cycle as injection.

Start with $1.$

$(34)$ does nothing to $1.$

$(12)$ takes $1$ to $2.$

$(123)$ takes $2$ to $3.$

So we write $(13$ and the rest of the cycle is still open ended.

what does $(123)(12)(34)$ do to $3?$

$(34)$ takes $3$ to $4.$ The rest of the cycles don't touch $4$

$(134$

What happens to $4?$

$(34)$ takes $4$ to $3, (12)$ does nothing, $(123)$ takes $3$ to $1$ This closes the cycle.

$(134)$

And we better hope that

$(123)(12)(34)$ does nothing to $2$ And as it turns out, it doesn't

Does this help?

One more example.

$(132)(12)(34)$

$1$ goes to $2$ which goes back to $1.$ We can write $(1)$ but it is not necessary

$2$ goes to $1$ which goes to $3$

$3$ goes to $4.$

$4$ goes to $3$ which goes to $2$

$(234)$

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    very helpful. I was told to start from the far right 1st but here you started left side first. I got $(123)(12)(32)$ correct by luck I guess. I adjusted my last part to $(132)(12)(34)$ to yours because I had $(214)$ because I thought the $2$ was being dropped – 2017-02-14
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    It is right to left. And that is what I did in my workings.. However, I did give you incorrect instructions. – 2017-02-14
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    Okay. for $(123)(34)$, I look at the $(34)$ 1st? If so then, I am getting $(234)$ – 2017-02-14
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    Regarding $(123)(34)$, start with $1$ look at what happens to $1$ when subjected to the $(34)$ cycle... nothing. Then the $(123)$ takes $1$ to $2.$ Then we look at what happens to $2....$ I get $(124)$ – 2017-02-14
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    thanks again. For some reason I had $(124)$ for my permutation $(132)(34)$. I starting to see it though. I still need more practice with this – 2017-02-14