Show that all proper subgroups of a group of order 15 are cy

Show that all proper subgroups of a group of order 15 are cyclic.

Solution

Let G be a group of order 15. We know G has subgroups of order 3 and order 5, say P3 and P5 from Sylow theory. These must be cyclic (why?) so write P3=a, P5=b.

Using the lemma below, show G=P3P5. Prove the lemma if it\'s not something you already know.

Lemma. For subgroups HH and KK of a finite group GG, |HK|=|H||K|/|HK|, where HK={hkhH,kK}.

Using Sylow theory, show P3 is normal.

Then bab^-1a. If bab^-1=a, we have ba=ab, so G is abelian. Observe bab^-11 (why?). The only \"bad\" possibility now is that bab^-1=a²

Suppose, to get a contradiction, that bab^-1=a². Then ba=a²b. Using this identity repeatedly to fill in the , show a=b⁵a==a²b⁵=a². But aa², so this is a contradiction.

PS - Since P3 and P5 are both normal, you could instead argue that G=P3P5 implies GP3×P5. In general, you can adapt this argument to show for primes p,q with p>q and qp1, every group of order pq is abelian.