MATH 433
April 17, 2015
Quiz 10: Solutions
Problem 1. We have the following information on a group G and its subgroups H1 and H2 :
• the order of G is either 15 or 20 or 25,
• H1 and H2 are proper subgroups of different orders and neither of them contains the other,
• the intersection H1 ∩ H2 is a nontrivial subgroup of G.
Find the order of H1 ∩ H2 .
Solution: 2.
Let n denote the order of the group G, let m1 and m2 denote orders of the subgroups H1 and H2 , and
let k denote the order of the intersection H1 ∩ H2 . We are given that n = 15 or 20 or 25, that m1 6= m2 ,
and that k > 1.
By Lagrange’s Theorem, m1 and m2 are proper divisors of n. Since neither of the subgroups H1 and
H2 contains the other, their intersection H1 ∩ H2 is a proper subgroup of H1 and of H2 . By Lagrange’s
Theorem, k is a proper divisor of m1 and of m2 . Since k > 1, it follows that neither m1 nor m2 is a prime
number or 1.
Now we obtain that n cannot be 15 or 25 as any proper divisor of these numbers is either prime or 1.
Therefore n = 20. The number 20 has only two proper divisors which are composite numbers: 4 and 10.
Hence one of them is m1 and the other is m2 . Finally, k is a common divisor of 4 and 10. Since k > 1, we
obtain that k = 2.
Remark. The group G in this problem is the dihedral group D(10), the group of symmetries of a regular
decagon.
Problem 2. Determine which of the following groups of order 4 are isomorphic and which are
not: Z4 , G5 , and G8 . Explain.
Solution: Z4 and G5 are isomorphic, the group G8 is not isomorphic to them.
The group Z4 is cyclic: Z4 = h[1]4 i. The group G5 = {[1]5 , [2]5 , [3]5 , [4]5 } is also cyclic: G5 =
{[1]5 , [2]5 , [4]5 , [8]5 } = h[2]5 i. Therefore Z4 is isomorphic to G5 . An isomorphism f : Z4 → G5 is given
by f ([n]4 ) = ([2]5 )n , n ∈ {0, 1, 2, 3}.
The group G8 = {[1]8 , [3]8 , [5]8 , [7]8 } is not cyclic. Indeed, [3]28 = [5]28 = [7]28 = [1]8 so that G8 has no
elements of order 4. It follows that G8 is not isomorphic to Z4 or to G5 .