Question

A group G of order 20 is __________

(a) solvable

(b) unsolvable

(c) 1

(d) not determined

I had been asked this question during an online interview.

My query is from Cyclic Groups in chapter Groups of Discrete Mathematics

Answer 1

The correct option is (a) solvable

The best I can explain: The prime factorization of 20 is 20=2⋅5. Let n5 be the number of 5-Sylow subgroups of G. By Sylow’s theorem, we have,  n5≡1(mod 5)and n5|4. Thus, we have n5=1. Let P be the unique 5-Sylow subgroup of G. The subgroup P is normal in G as it is the unique 5-Sylow subgroup. Then consider the subnormal series G▹P▹{e}, where e is the identity element of G. Then the factor groups G/P, P/{e} have order 4 and 5 respectively. Hence these are cyclic groups(in particular abelian). Hence, the group G of order 20 has a subnormal series whose factor groups are abelian groups, and thus G is a solvable group.