Counting Cosets: Clarifying Right & Left Cosets

[onie mti]

i am reading a chapter on counting cosets and I am not sure i fully understand the theory behind right and left cosets. can i please be given clear descriptions perhaps with examples.

 

[Fermat1]

Re: counting cosets

i am reading a chapter on counting cosets and I am not sure i fully understand the theory behind right and left cosets. can i please be given clear descriptions perhaps with examples.

You should post what you are specifically having difficulties with. Maybe give a problem which you can't solve.

 

[onie mti]

Re: counting cosets

You should post what you are specifically having difficulties with. Maybe give a problem which you can't solve.

i am given that H is a subgp of G, list the coset of H, for each coset list the elements of the coset

G=s_3, H= {epsilon, beta, alpha}

 

[Fermat1]

Re: counting cosets

i am given that H is a subgp of G, list the coset of H, for each coset list the elements of the coset

G=s_3, H= {epsilon, beta, alpha}

I'm not sure on your epsilon, beta, alpha notation. $S_{3}$ is the set of permutations of the vector (1,2,3).

 

[Deveno]

Let's pick a specific group, so we can be definite about this. To make things easier, we'll pick a rather small group.

Specifically, let:

$G = \{e,a,a^2,b,ab,a^2b\}$ where:

$a^3 = b^2 = e$, and the multiplication is given by the rule:

$ba = a^2b$.

Now we need to pick a subgroup, so let $H$ be the subgroup:

$H = \{e,b\}$.

Let's look at the left cosets first:

$eH = H = \{e,b\}$.
$aH = \{a,ab\}$
$a^2H = \{a^2,a^2b\}$
$bH = \{b,e\} = H$
$abH = \{ab,a\} = aH$
$a^2bH = \{a^2b,a^2\} = a^2H$

so we have 3 different (distinct) left cosets: $H,aH,a^2H$.

Now, let's look at the right cosets:

$He = H = \{e,b\}$
$Ha = \{a,ba\} = \{a,a^2b\}$ (see above)
$Ha^2 = \{a^2,ba^2\} = \{a^2,ab\}$

(because $ba^2 = (ba)a = (a^2b)a = a^2(ba) = a^2(a^2b) = a^4b = (a^3)(ab) = ab$).

$Hb = \{b,e\} = H$
$Hab = \{ab,bab\} = \{ab,a^2\} = Ha^2$
$Ha^2b = \{a^2b,ba^2b\} = \{a^2b,a\} = Ha$

Again, we have three distinct right cosets, as well: $H,Ha,Ha^2$.

But notice these are DIFFERENT sets than the left cosets, for while:

$eH = H = He$
$bH = H = Hb$

we have:

$aH = \{a,ab\}$ which has $ab$ as a member, whereas:

$Ha = \{a,a^2b\}$, which does not have $ab$ as a member.

Now this is bad news if we were planning on trying to define:

$(Ha)(Hb) = Hab$.

Let's look at what the set $(Ha)(Hb) = \{xy: x\in Ha,y \in Hb\}$ actually works out to be. We have 4 possible products:

$ae,ab,(a^2b)(e),(a^2b)(b)$, and working out what these are gives us:

$ae = a$
$ab = ab$
$a^2b = a^2b$
$(a^2b)(b) = a^2$

so $(Ha)(Hb) = \{a,a^2,ab,a^2b\}$ and this isn't even ANY coset of $H$ (all the cosets of $H$ have 2 elements, the same number of elements $H$ has, but this product set has FOUR elements).