Problem 4 (b) in the assignment: hint

It suffices to set up the bijection by explaining what it does to elements in G/K. For example, the coset of 1  in the factor group G/K is 1+K={1,10,19,28}. The coset of 1 in the factor group G/H is 1+H={1,19}. The bijection in the third isomorphism theorem pairs the coset (1+H)+K/H  in the factor group (G/H)/(K/H) to the coset 1+K in G/K. Concretely, K/H has the two cosets H and K\setminusH={9,27}, so (1+H)+K/H ={{1,19},{10,28}}={1+H, 10+H}. Thus the bijection pairs {{1,19},{10,28}} with {1,10,19,28}. Similarly for 0+K,2+K,..., 8+K in G/K.