Compute the order of the element r2 in the factor group D6So

Compute the order of the element r^2<r^3> in the factor group D6/<r^3>.

Solution

D₆ is the dihedral group of order 12.

D₆ = {e, r, r², r³, r⁴, r⁵, f₁, f₂, f_3, f₄, f₅, f₆}

Where r represent rotation of a regular hexagon by 60⁰, f_i represent flip with respect a line of symmetry (i = 1, 2, 3, 4, 5, 6). There are total 6 lines of symmetries for a regular hexagon. Hence, there are 6 rotations and 6 flips. Note that r⁶ = e = f_i² for all i = 1, 2, 3, 4, 5, 6.

D₆ has two Normal subgroups H and K given as H = <r> = {e, r, r², r³, r⁴, r⁵}, K = <r³> = {e, r³}

H is normal subgroup of D₆ because the index of H in D₆ is |D₆ : H|=|D₆|/|H| = 12/6 = 2 (Recall the property that if a subgroup H of a finite group G has index |G:H| = 2, then H is normal subgroup of G).

K is normal subgroup of D₆ because K = Z(D₆) i.e. K is the center of D₆. Recall the property that center Z(G) of a group G is a normal subgroup of G.

Hence, the factor group D₆/K is defined. The order of D₆/K is |D₆/K| = |D₆|/|K| = 12/2 = 6

D₆/K = {xK : x D₆}

The elements of D₆/K are left cosets of K in D₆. The identity element of D₆/K is eK and the non-identity elements of D₆/K are of the form xK where x D₆ – K.

Order of a non-identity element xK in D₆/K is equal to order of the element x in D₆. This is easy to prove. Let order of x in D₆ is n. Then xⁿ = e. Now, (xK)ⁿ = xⁿK = eK. This implies that order of xK in D₆/K is n. Now, order of xK in D₆/K cannot be less than n. Let if possible order of xK in D₆/K is m and m < n. Then (xK)^m = eK, but (xK)^m = x^mK. This implies x^mK = eK, i.e. x^m = e, a contradiction to the assumption that order of x in D₆ is n. Therefore, we have proved that order of xK in D₆/K is n.

Hence, order of r²K in D₆/K is equal to the order of the element r² in D₆. Order of r² in D₆ is 3 as (r²)³ = r⁶ = e. Therefore, order of r²K in D₆/K is equal to 3. (Answer)