 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-2/8/ii/solution

H

G

∣ G ∣ = [G : H] ∣ H ∣.

The left cosets partition the group and each has

∣ H ∣

members, proving the formula. If

∣ G ∣ = p

is prime, the only divisors are one and

p

, so

H = {e}

H = G

. Hence a group of prime order has no nontrivial proper subgroups.

The word “nontrivial” is necessary:

{e}

is a proper subgroup of every prime-order group. Thus the source's shorter wording cannot literally exclude all proper subgroups.

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