Find the order of the following
(π2 Γ π4)/< (1,1) >
(1,1)+(1,1)=(0,2)(0,2)+(1,1)=(1,3)(1,3)+(1,1)=(0,0)fromββwhichβ£<(1,1)>β£=β£{(1,1),(0,2),(1,3),(0,0)}β£=4β£(Z2ΓZ4)/<(1,1)>β£=2β 44=2\left( 1,1 \right) +\left( 1,1 \right) =\left( 0,2 \right) \\\left( 0,2 \right) +\left( 1,1 \right) =\left( 1,3 \right) \\\left( 1,3 \right) +\left( 1,1 \right) =\left( 0,0 \right) \\from\,\,which\\\left| <\left( 1,1 \right) > \right|=\left| \left\{ \left( 1,1 \right) ,\left( 0,2 \right) ,\left( 1,3 \right) ,\left( 0,0 \right) \right\} \right|=4\\\left| \left( Z_2\times Z_4 \right) /<\left( 1,1 \right) > \right|=\frac{2\cdot 4}{4}=2(1,1)+(1,1)=(0,2)(0,2)+(1,1)=(1,3)(1,3)+(1,1)=(0,0)fromwhichβ£<(1,1)>β£=β£{(1,1),(0,2),(1,3),(0,0)}β£=4β£(Z2βΓZ4β)/<(1,1)>β£=42β 4β=2
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