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