Thread: Subgroup coset

How can i find left and right cosets of subgroup <2> ={0,2,4,6,8,10} of Z₁₂ ???
thanks

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

Z₁₂ is abelian, so the left cosets and right cosets are the same.

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.

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 ??

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.