# Subgroup coset

• Jun 17th 2012, 03:33 AM
kljoki
[SOLVED]Subgroup coset
How can i find left and right cosets of subgroup <2> ={0,2,4,6,8,10} of Z12 ???
thanks
• Jun 17th 2012, 12:01 PM
kljoki
Re: Subgroup coset
there is 2 cosets for H={0,2,4,6,8,10}
0+H = H
1+H = {1,3,5,7,9,11}
next is the same with this 2 :)
• Jun 17th 2012, 12:40 PM
Deveno
Re: Subgroup coset
Z12 is abelian, so the left cosets and right cosets are the same.
• Jun 17th 2012, 05:22 PM
HallsofIvy
Re: Subgroup coset
And there are two such cosets because the given sugroup contains 6= 12/2 members. kljoki, this would have been a good exercise in using the basic definition of "coset" because that is all that is required.
• Jun 18th 2012, 04:10 AM
kljoki
Re: Subgroup coset
Quote:

Originally Posted by Deveno
Z12 is abelian, so the left cosets and right cosets are the same.

and what if it wasn't abelian, can you give me some example ?? :D
• Jun 18th 2012, 05:42 AM
Deveno
Re: Subgroup coset
sure. let's use the group of symmetries of a triangle. this is generated by a rotation of 120 degrees, and a reflection about the x-axis, which i shall call a and b. the multiplication operation is composition of functions (we consider the domain of these functions to be the triangle itself; although if you wish, we can consider them as acting on the entire euclidean plane).

clearly, a3 = e, the identity function ("do nothing" = "rotating 360 degrees")

also clearly, b2 = e.

by actual experimentation (you could make a triangle out of straws, for example), we see that:

ba = a2b.

thus G = {e,a,a2,b,ab,a2b}.

we have an obvious subgroup of order 2: H = {e,b}. let's compute the left and right cosets of H. first the left cosets:

eH = {ee,eb} = {e,b} = H (this one is easy).
aH = {ae,ab} = {a,ab}
a2H = {a2e, a2b} = {a2,a2b}

so far, so good. these 3 cosets are clearly all different.

bH = {be,bb} = {b,b2} = {b,e} = H (remember, in SETS, order of elements doesn't matter).
(ab)H = {(ab)e, (ab)b} = {ab,ab2} = {ab,a} = aH (not surprising since (ab)H should equal a(bH) = aH, since bH = H).
(a2b)H = {(a2b)e, (a2b)b} = {a2b, a2} = a2H

so we only have 3 "distinct" left cosets, H, aH and a2H.

now let's look at the right cosets of H:

He = H (this really should be obvious).
Ha = {ea,ba} = {a,a2b} (remember that ba = a2b)
Ha2 = {ea2,ba2}.

hmm. what *is* ba2? we don't rightly know yet. let's figure it out:

ba2 = (ba)a = (a2b)a = a2(ba) = a2(a2b) = a4b = (a3)(ab) = (e)(ab) = ab.

so Ha2 = {a2, ab}.

note that aH is a different set than Ha, and a2H is a different set than Ha2. this shows that when G is non-abelian, for some subgroups H it might be possible that gH ≠ Hg.

i leave it to you to verify the the rest of the cosets are:

Hb = H
H(ab) = Ha2
H(a2b) = Ha

you will discover (shortly) that subgroups for which the left and right cosets are the same, are "special", and have their own name: normal subgroups.