Question Number 157438 by apriadodir last updated on 23/Oct/21
$$\mathrm{find}\:\mathrm{all}\:\mathrm{subgroups}\:\mathrm{of}\:: \\ $$$$\left.\mathrm{a}\right)\:\mathrm{grup}\:\left(\mathrm{Z}_{\mathrm{6}} \:,\:+\right) \\ $$$$\left.\mathrm{b}\right)\:\mathrm{grup}\:\left(\mathrm{Z}_{\mathrm{6}} \:−\left\{\mathrm{0}\right\},\:×\right) \\ $$
Answered by mindispower last updated on 23/Oct/21
![if G is a subgroup of E card(G)∣card(E)⇒card(G)∈{1,2,3} 0∈G card(G), since (Z_6_, ,+) is Cycl groupe (Z_6 ,+)=<a> withe a 6=1,a∈F={1,3,5} if a=2, A=<2>,card(A)=3, a=3,A=<3> card(A)=2 a=4,card(A)=3,A=0 card(A)=1 a∈F (z_6 ,+)=<a>=card =6 b) card grup (z_6 −{0},∗) is not a grup (2),(3)∈ but 2∗3=6=0[6]∉](Q157445.png)
$${if}\:{G}\:{is}\:{a}\:{subgroup}\:{of}\:{E} \\ $$$${card}\left({G}\right)\mid{card}\left({E}\right)\Rightarrow{card}\left({G}\right)\in\left\{\mathrm{1},\mathrm{2},\mathrm{3}\right\} \\ $$$$\mathrm{0}\in{G}\:\:\:{card}\left({G}\right), \\ $$$${since}\:\left({Z}_{\mathrm{6}_{,} } ,+\right)\:{is}\:{Cycl}\:{groupe} \\ $$$$\left({Z}_{\mathrm{6}} ,+\right)=<{a}>\:\:\:{withe}\:{a} \mathrm{6}=\mathrm{1},{a}\in{F}=\left\{\mathrm{1},\mathrm{3},\mathrm{5}\right\} \\ $$$${if}\:{a}=\mathrm{2},\:{A}=<\mathrm{2}>,{card}\left({A}\right)=\mathrm{3}, \\ $$$${a}=\mathrm{3},{A}=<\mathrm{3}>\:{card}\left({A}\right)=\mathrm{2} \\ $$$${a}=\mathrm{4},{card}\left({A}\right)=\mathrm{3},{A}=\mathrm{0}\:{card}\left({A}\right)=\mathrm{1} \\ $$$${a}\in{F}\:\:\:\left({z}_{\mathrm{6}} ,+\right)=<{a}>={card}\:=\mathrm{6} \\ $$$$\left.{b}\right)\:{card}\:{grup}\:\left({z}_{\mathrm{6}} −\left\{\mathrm{0}\right\},\ast\right)\:{is}\:{not}\:{a}\:{grup} \\ $$$$\left(\mathrm{2}\right),\left(\mathrm{3}\right)\in\:{but}\:\mathrm{2}\ast\mathrm{3}=\mathrm{6}=\mathrm{0}\left[\mathrm{6}\right]\notin \\ $$$$ \\ $$