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Here's the problem in Chapter 7 of Gallian's Contemporary Abstract Algebra (not homework, just furthering my understanding): Let $H = \{(1), (12)(34), (13)(24), (14),(23)\}$. Find the left cosets of $H$ in $A_4$.

The answer provided in the back of the book is

$$H = \{\alpha_1, \alpha_2, \alpha_3, \alpha_4\}$$ $$\alpha_5 H = \{\alpha_5, \alpha_8, \alpha_6, \alpha_7\}$$ $$\alpha_9 H = \{\alpha_9, \alpha_{11}, \alpha_{12}, \alpha_{10}\}$$

My primary hang up is the abstract nature of the solution. The other examples in the book are fairly concrete in that they deal with specific numbers so I'm not sure why this doesn't.

I thought a coset meant for each element in $A_4$ apply the group operation for every element in $H$. I see that that appears to be happening but only in increments of 4 in $A_4$ (ex. $a_1, a_5, a_9, ...$).

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migrated from mathoverflow.net May 12 '14 at 13:11

This question came from our site for professional mathematicians.

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You're right that there is a coset for every element in $A_4$, but lots of them coincide: here $H = \alpha_1 H = \alpha_2 H = \alpha_3 H = \alpha_4 H$, $\alpha_5 H = \alpha_6 H = \alpha_7 H = \alpha_8 H$, and $\alpha_9 H = \alpha_{10} H = \alpha_{11} H = \alpha_{12} H$. It seems that the written answer decided to include each coset only once, which is perfectly reasonable.

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  • $\begingroup$ Oh of course! Thank you very much. $\endgroup$ – Justin Bozonier May 12 '14 at 13:17

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