Let \(G\) be a group of 35 elements. Then the largest possible size of a…

Answer: 7. The largest proper subgroup has order 7.

Explanation: By Lagrange's theorem, the order of any subgroup must divide the group order 35, so the possible subgroup orders are 1, 5, 7, and 35. Thus the largest proper divisor is 7, but we must also show a subgroup of order 7 actually exists.

• Apply Sylow's theorem for p = 7: the number n_7 of Sylow-7 subgroups satisfies n_7 ≡ 1 (mod 7) and n_7 divides 5, hence n_7 = 1. Therefore a subgroup of order 7 exists (in fact it is unique and thus normal).

• For completeness, Sylow's theorem for p = 5 gives n_5 ≡ 1 (mod 5) and n_5 divides 7, so n_5 = 1, so a subgroup of order 5 also exists but is smaller than 7.

Hence the largest possible size of a subgroup of G other than G itself is 7.