Cyclic Subgroups: Finite Groups

Date: 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