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### 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
```
Associated Topics:
College Modern Algebra

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