Cyclic Subgroups: Finite GroupsDate: 02/01/2002 at 20:26:11 From: Dani Subject: Cyclic subgroups: Finite Groups Is there a noncyclic subgroup of order 4 in U(40)? If so how can it be found? Sincerely, Dani Date: 02/02/2002 at 00:04:13 From: Doctor Paul Subject: Re: Cyclic subgroups: Finite Groups There are two isomorphism classes for groups of order four: either a group of order four is isomorphic to the cyclic group Z_4 or a group of order four is isomorphic to the Klein Four group. Recall that the Klein Four group is noncyclic since every element squared is the identity (i.e., every nonidentity element has order two). This means that the group has no element of order four and hence cannot be cyclic. So if there is a noncyclic subgroup of order 4 in U(40) = {x in Z_40 | gcd(x,40) = 1} then it must be isomorphic to the Klein Four group. So it must have the identity element and it must have three elements whose square is the identity. In other words, you want to find elements x in U(40) such that x^2 is the identity element (i.e., 1) in U(40). Recall that multiplication is reduced mod 40 in the group U(40). So asking that x^2 be the identity element in U(40) is equivalent to asking that x^2 = 1 mod 40. To find such values for x, I just started looking at numbers that were congruent to 1 mod 40 and checked to see if they were perfect squares. 1 is a perfect square so sqrt(1) = 1 must be in our noncyclic subgroup of order four. 41 isn't a perfect square. 81 is a perfect square so sqrt(81) = 9 must be in our noncyclic subgroup of order four. 121 is a perfect square so sqrt(121) = 11 must be in our noncyclic subgroup of order four. So far we have {1, 9, 11, __ } Since we know that this is a subgroup, it must be closed under the group operation (multiplication), so 9*11 = 99 = 19 must be the fourth element in the group. To verify this, we check that 19^2 = 361 = 1 mod 40. Thus we have the noncyclic subgroup {1, 9, 11, 19} with multiplication table as follows: * | 1 9 11 19 ----------------------- 1 | 1 9 11 19 | 9 | 9 1 19 11 | 11 | 11 19 1 9 | 19 | 19 11 9 1 As desired, this group is isomorphic to the Klein Four group. I hope this helps. Please write back if you'd like to talk about this more. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/ Date: 02/02/2002 at 09:25:07 From: Dani Subject: Cyclic subgroups: Finite Groups Thank you for sharing your knowledge. It is quite easy to see the process from your steps. I appreciate your taking the time to help me with what looks like a rather simple problem now. :) Thank you, Dani |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/