# coset multiplication when $H = \{(1),(12)(34)\}$

Let $H = \{(1),(12)(34)\}$ and $\alpha_1 = (243), \alpha_2 = (142), \alpha_3 = (132),$ and $\alpha_4 = (234)$.

Coset multiplication is a bit confusing to me. The book states in an example that $\alpha_1H = \alpha_2H$ and I'd like to see it for myself, however not a single paragraph in this book actually explains the operation of $\alpha_1H$. As to why I wrote the other $\alpha_i$'s, the book claims that:

$\alpha_1 \alpha_3H \neq \alpha_2 \alpha_4H$

I just would like to see it for myself before I press forward.

Thanks!

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$\alpha_1 H$ just means you take every element in $H$ and multiply on the left by $\alpha_1$. So $$\alpha_1 H = \{\alpha_1 (1), \alpha_1 (12)(34)\} = \{(243)(1), (243)(12)(34)\} = \{(243), (142)\}$$ Can you compute $\alpha_2 H$ yourself? What about $\alpha_1\alpha_3 H$ and $\alpha_2 \alpha_4 H$?