Are the Right and Left Cosets Equal in a Group's Cayley Table?

[mathsdespair]

Just by looking at the cayley table of a group and looking at its subgroups, is their a theorem or something which tells you if the right and left cosets are equal?

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

 

[homeomorphic]

http://en.wikipedia.org/wiki/Normal_subgroup

 

[mathsdespair]

http://en.wikipedia.org/wiki/Normal_subgroup

ok thanks

 

[mathsdespair]

When multiplying cosets, can you just quite simply multiply them together?
Our teacher said something about their could be a potential problem?
What could the problem be?

 

[jbunniii]

When multiplying cosets, can you just quite simply multiply them together?
Our teacher said something about their could be a potential problem?
What could the problem be?

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

 

[mathsdespair]

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

Thank you for the good explanation.