Then there must be two elements in the order 15. Elements a and b are all of order 5, and A N and B M are not equal to all m and n. ..

So ab is not here

& lta>,<b>, {ab, ab2, ab3, ab4}, {a2b, a3b, a4b, a5b} must all be in the original group.

But there are three more elements in a * * *, so there must be some elements equal in the last two sets, because they can't be equal to the previous < a> and.

A = b (-p-q), then

That is, any two cyclic groups of order 5.