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Product of a Finite Abelian Group

```Date: 06/28/2004 at 11:19:21
From: Angie
Subject: What can I say about the "product" of a finite Abelian group

Suppose G = {a1, a2, ... , an} is a finite Abelian group.  If G has
odd order, what can you say about the "product," a1*a2*...*an, of all
the elements of G?  What can you say about this "product" if G has
even order?  What if G is not Abelian?

I am not sure how * not being commutative affects the "product" of the
elements of the group.

My thoughts:

If G has odd order, then it has an odd number of elements one of which
is the identity and the rest would be elements and their inverses.

e*(a1*a2*a3*...*an*a1'*a2'*a3'*...*an')--->

e*(a1*a1'*a2*a2'*...*an*an')--->

e*e = e

But what if some of the elements are their own inverses?

e*(a1*a2*a3*a4*...*an*a3'*a4'*...*an') --->

e*(a1*a2)*(a3*a3'*a4*a4'*...*an*an')--->

e*(a1*a2)*e ---> e*a1*a2 ---> e*a ---> a

where a1*a2 = a, which would be an element of the group G (G is closed
under *).

If G has even order, then it has an even number of elements one of
which is the identity, e, and then an odd number of elements.

There would have to be an odd number of elements that are there own
inverses and an even number of elements with an even number of
inverses?

e*(a1*a2*a3*a4*a5*...*an*a4'*a5'*...*an') --->

e*(a1*a2*a3*a4*a4'*a5*a5'*...*an*an') --->

e*(a1*a2*a3)*(a4*a4'*a5*a5'*...*an*an') --->

e*(a1*a2*a3)*e ---> e*(a1*a2*a3) ---> e*a ---> a

where a1*a2*a3 = a and a is an element of G since G is closed under *.

```

```
Date: 06/28/2004 at 20:51:20
From: Doctor Vogler
Subject: Re: What can I say about the "product" of a finite Abelian group

Hi Angie,

We got a question very similar to yours a short time ago, so I've
thought about this before.  It sounds like you might be forgetting
that your group is not necessarily cyclic.  But it is close.

The Fundamental Theorem of Finitely Generated Abelian Groups says that
all finite Abelian groups are a product of cyclic groups.  Use that
fact, and then it might help to write it out additively instead of
multiplicatively.

Suppose that our group is (Z/aZ) x (Z/bZ) x (Z/cZ).  How do we find
the sum of all of the a*b*c elements of this group?  Remember that we
add termwise, so we have

a-1 b-1 c-1                   a-1        b-1        c-1
sum sum sum (r, s, t) = ( b*c*sum r, a*c*sum s, a*b*sum t )
r=0 s=0 t=0                   r=0        s=0        t=0

= (b*c*a*(a-1)/2, a*c*b*(b-1)/2, a*b*c*(c-1)/2).

Now the first term is considered mod a, the second mod b, the third
mod c.  At first glance, you think: "a divides the first term, b the
second, and c the third, so the sum is the identity (0, 0, 0)," but
that's not quite true; you have to divide by 2 first.  If a, b, or c
is odd, then 2 has an inverse, and so the corresponding term *is*
zero.  If not, then you have to divide one of the even numbers in the
product by 2 first.  Suppose that two of a, b, and c are even.  Then
each term will still be zero, because you can divide the two out of
the other number.  But if only one of a, b, or c is even (say a), then
you get

(b*c*(a/2)*(a-1), (a/2)*c*b*(b-1), (a/2)*b*c*(c-1))
= (b*c*(a/2)*(a-1), 0, 0)

since the last two terms are divisible by b and c, respectively.
Furthermore, (a/2)*k mod a is determined by whether k is odd or even,
since 2*(a/2) = 0 mod a, and since b, c, and a-1 are all odd, we have

(a/2, 0, 0).

Since a/2 is the unique element of order 2 in Z/aZ, and the other
component groups have odd order, this is the unique element of order 2
in our Abelian group, that is, the element -1 when the group is
considered multiplicatively.

I used three cyclic components to have a concrete example to use, but
the argument does not depend on the three.  You can generalize this to
any finite Abelian group, and the result is the same:  If there is
exactly one cyclic component of even order, then the sum (product) of
all of the elements is the unique element of order 2.  Otherwise, the
sum (product) of all of the elements is the identity.  In particular,
the cyclic group of order 8 is different from the product of three
cyclic groups of order 2 (which group also has order 8).  In the
second group, the sum (product) of all 8 elements is the identity, but
not in the first.

For a concrete example, let us examine the product of all elements
less than n and relatively prime to n in the finite Abelian group
(Z/nZ)*, the multiplicative group of invertible elements mod n.
Another well-known theorem from abstract algebra tells exactly how
this group is structured:

If n = r*s and r and s are relatively prime, then we have the
isomorphism

(Z/nZ)* = (Z/rZ)* x (Z/sZ)*.

If n = p^k for a prime p and k>=1, then we have the isomorphism

(Z/nZ)* = (Z/(p-1)Z) x (Z/(p^(k-1))Z) for odd p,
(Z/nZ)* = (Z/2Z) x (Z/(2^(k-2))Z) for p=2 and k>1, and
(Z/2Z)* = 1 (the trivial group).

So you can use this to decompose any group.  For example,

(Z/54Z)* = (Z/2Z)* x (Z/27Z)*
= 1 x (Z/2Z) x (Z/9Z)
= (Z/2Z) x (Z/9Z),

which has one component of even order, so the product of the elements
is -1.  In fact, using all of these ideas, it is not hard to show that
the product of the elements of (Z/nZ)* is -1 if and only if n is:

2 (in which case 1 = -1),
4,
p^k (for some odd prime p), or
2*p^k (again, for an odd prime p).

In the case n = 2 and all other cases, the product is 1.

Finally, you asked about non-Abelian groups.  For this, I might ask
you exactly what you mean by the product of all of the elements.  For
example, in the non-Abelian group S3, of order 6, you might mean

1 (12) (23) (13) (123) (132)

or you might mean

1 (123) (12) (13) (132) (23)

or any of a variety of other things.  Do you always get the same
product?  Do you when the group is Abelian?

If you have any questions or need more help, please write back and
show me what you have been able to do, and I will try to offer further
suggestions.

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 06/28/2004 at 21:00:00
From: Angie
Subject: Thank you (What can I say about the "product" of a finite
Abelian group)

Thank you, Dr. Vogler.  I think that the "product" would be the same
no matter what the order of the elements.

I don't understand how not being able to change the order in which
the operation acts on the elements would make any difference in the
final product.

```

```
Date: 06/29/2004 at 17:13:42
From: Doctor Vogler
Subject: Re: What can I say about the "product" of a finite Abelian group

Hi Angie,

In my noncommutative example, I mentioned

1 (12) (23) (13) (123) (132)

and

1 (123) (12) (13) (132) (23).

The first product is (32).  The second is (13).  The order changes the
value of the product.

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 06/29/2004 at 18:14:40
From: Angie
Subject: Thank you (What can I say about the "product" of a finite
Abelian group)

Thank you.  I have to remember that the "operation" is not just
multiplication.

Thanks again.
```
Associated Topics:
College Modern Algebra

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