Solution: Given group $G=({Z}_{15},\oplus )$ a) We know that by Lagrange`s theorem order of subgroup divide the order of group. Since $O\left(G\right)=15$ Let H be subgroup of G. Then $O\left(H\right)|O\left(G\right)\cdot O\left(H\right)|15$ Possible order of H are 1, 3, 5, 15 Since $G-({Z}_{15},\oplus )$ is cycling group. Then every divisor of order of group has subgroup. Then b)

${H}_{1}=\left\{e\right\}$

${H}_{2}=<5>=\{5,10,0\}$

${H}_{3}=<5>\{3,6,9,12,0\}$

${H}_{4}=<1>=\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,0\}$

These are $H}_{1},{H}_{2},{H}_{3}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}{H}_{4$ are subgroup of group $G=({Z}_{15},\oplus )$