### 30th Putnam 1969

**Problem B2**

G is a finite group with identity 1. Show that we cannot find two proper subgroups A and B (≠ {1} or G) such that A ∪ B = G. Can we find three proper subgroups A, B, C such that A ∪ B ∪ C = G?

**Solution**

We have |A| divides |G| and hence |A| ≤ |G|/2. Similarly, |B| ≤ |G|/2. But 1 belongs to both A and B, so |A union B| < |G|.

For three subgroups, sometimes we can and sometimes we cannot. For example, if G is the group of order 4 defined by ba = ab, a^{2} = b^{2} = 1, then we can: A = {1, a}, B = {1, b}, C = {1, ab}. On the other hand, if G is a cyclic group of prime order, then it has no proper subgroups. [But these are far from the only exceptions. For example take G to be cyclic of order 15. Then any proper subgroup must have order 3 or 5. All subgroups contain 1, so any three subgroups contain at most 13 elements between them.]

30th Putnam 1969

© John Scholes

jscholes@kalva.demon.co.uk

14 Jan 2002