- #1

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I have question to do and I would love to half the workload by not having to to work out the same thing twice.

Thanks

- Thread starter mathsdespair
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- #1

- 16

- 0

I have question to do and I would love to half the workload by not having to to work out the same thing twice.

Thanks

- #2

- #3

- #4

- 16

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Our teacher said something about their could be a potential problem?

What could the problem be?

- #5

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You can always multiply cosets, but the result is not necessarily a coset. In other words, the set of right or left cosets is not generally a group. Thus you can form the product ##aHbH##, which is the set of all elements of the form ##ah_1bh_2##, but this does not generally equal ##abH##, nor can it generally be written in the form ##gH## at all.

Our teacher said something about their could be a potential problem?

What could the problem be?

However, if ##H## is a normal subgroup, then the left and right cosets are the same (##aH = Ha##), so we just call them cosets, and the set of cosets does form a group. In this case, the product ##aHbH## can be simplified as follows:

$$aHbH = a(Hb)H = a(bH)H = abHH = abH$$

- #6

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Thank you for the good explanation.You can always multiply cosets, but the result is not necessarily a coset. In other words, the set of right or left cosets is not generally a group. Thus you can form the product ##aHbH##, which is the set of all elements of the form ##ah_1bh_2##, but this does not generally equal ##abH##, nor can it generally be written in the form ##gH## at all.

However, if ##H## is a normal subgroup, then the left and right cosets are the same (##aH = Ha##), so we just call them cosets, and the set of cosets does form a group. In this case, the product ##aHbH## can be simplified as follows:

$$aHbH = a(Hb)H = a(bH)H = abHH = abH$$

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