How can i find left and right cosets of subgroup <2> ={0,2,4,6,8,10} of Z_{12}???

thanks

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- June 17th 2012, 04:33 AMkljoki[SOLVED]Subgroup coset
How can i find left and right cosets of subgroup <2> ={0,2,4,6,8,10} of Z

_{12}???

thanks - June 17th 2012, 01:01 PMkljokiRe: 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 :) - June 17th 2012, 01:40 PMDevenoRe: Subgroup coset
Z

_{12}is abelian, so the left cosets and right cosets are the same. - June 17th 2012, 06:22 PMHallsofIvyRe: 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. - June 18th 2012, 05:10 AMkljokiRe: Subgroup coset
- June 18th 2012, 06:42 AMDevenoRe: 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, a^{3}= e, the identity function ("do nothing" = "rotating 360 degrees")

also clearly, b^{2}= e.

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

ba = a^{2}b.

thus G = {e,a,a^{2},b,ab,a^{2}b}.

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}

a^{2}H = {a^{2}e, a^{2}b} = {a^{2},a^{2}b}

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

bH = {be,bb} = {b,b^{2}} = {b,e} = H (remember, in SETS, order of elements doesn't matter).

(ab)H = {(ab)e, (ab)b} = {ab,ab^{2}} = {ab,a} = aH (not surprising since (ab)H should equal a(bH) = aH, since bH = H).

(a^{2}b)H = {(a^{2}b)e, (a^{2}b)b} = {a^{2}b, a^{2}} = a^{2}H

so we only have 3 "distinct" left cosets, H, aH and a^{2}H.

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

He = H (this really should be obvious).

Ha = {ea,ba} = {a,a^{2}b} (remember that ba = a^{2}b)

Ha^{2}= {ea^{2},ba^{2}}.

hmm. what *is* ba^{2}? we don't rightly know yet. let's figure it out:

ba^{2}= (ba)a = (a^{2}b)a = a^{2}(ba) = a^{2}(a^{2}b) = a^{4}b = (a^{3})(ab) = (e)(ab) = ab.

so Ha^{2}= {a^{2}, ab}.

note that aH is a different set than Ha, and a^{2}H is a different set than Ha^{2}. 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) = Ha^{2}

H(a^{2}b) = 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.