# Quick help

• Nov 12th 2009, 10:15 AM
biggybarks
Quick help
Is this right.

Suppose that G is an Abelian Group with an odd number of elements.?

Show that the product of all of the elements of G is the identity

For ease, let's just show Abelian G with 7 elements. g = {e, a, b, c, d, f, g} --> e is the identity element.

Now consider a*b*c*d*e*f*g. Since the group is finite and abelian, then one of {b, c, d, f, g} must be the multiplicative inverse of a. Without loss of generality, we can presume that d = a^-1, f = b^-1 and g = c^-1.

a*b*c*d*e*f*g = a*d*b*f*e*c*g (commutative property)
= (a*d)*(b*f)*e*(c*g) (associative property)
= e*e*e*e (since d = a^-1, f = b^-1 and g = c^-1)
= e
• Nov 12th 2009, 10:22 AM
tonio
Quote:

Originally Posted by biggybarks
Is this right.

Suppose that G is an Abelian Group with an odd number of elements.?

Show that the product of all of the elements of G is the identity

For ease, let's just show Abelian G with 7 elements. g = {e, a, b, c, d, f, g} --> e is the identity element.

Now consider a*b*c*d*e*f*g. Since the group is finite and abelian, then one of {b, c, d, f, g} must be the multiplicative inverse of a. Without loss of generality, we can presume that d = a^-1, f = b^-1 and g = c^-1.

a*b*c*d*e*f*g = a*d*b*f*e*c*g (commutative property)
= (a*d)*(b*f)*e*(c*g) (associative property)
= e*e*e*e (since d = a^-1, f = b^-1 and g = c^-1)
= e

Very nice! Now just generalize to any odd number of elements.

Tonio
• Nov 12th 2009, 10:26 AM
signaldoc
You have implicitly assumed that no element is it's own inverse. This is true, by Lagrange, but should probably be included in your argument.
• Nov 12th 2009, 10:31 AM
biggybarks
This can be generalized to any Abelian group G with an odd number of elements; the process is the same. Find the multiplicative inverses, commute them together, then begin cancelling identities until you are left only with e.

How bout this. would this finish ir off or am I still missing something
• Nov 12th 2009, 10:41 AM
signaldoc
Yes, you have it. The argument is like this: The group is finite. No element is its own invrse. In the product of all elements, group inverse pairs together (to form the identity). This process ends after a finite number of steps with a single unpaired element, the identity, remaining. The product of a finite number of identity elements is the identity.
• Nov 12th 2009, 10:47 AM
tonio
Quote:

Originally Posted by signaldoc
Yes, you have it. The argument is like this: The group is finite. No element is its own invrse.

We know...but why?

Tonio

In the product of all elements, group inverse pairs together (to form the identity). This process ends after a finite number of steps with a single unpaired element, the identity, remaining. The product of a finite number of identity elements is the identity.

.
• Nov 12th 2009, 10:48 AM
Drexel28
Yeah. Here would be a formal proof

Problem: Let $\left\langle G,\circ\right\rangle$ be a group such that $|G|=n$ be odd. Compute $\prod_{g\in G}g$.

Proof: Note that $a\ne a^{-1}$ for any $a\in G$ otherwise $|a|=2$, which clearly cannot be the case since $|G|$ is odd and $|a|\text{ }|\text{ }|G|$. Then, $\forall g\in G$ there exists some $g'\ne g$ such that $g'=g^{-1}$. Thus pair off all the elements of $G-\{e_G\}$ as $\left\{g,g^{-1},g_1,g_1^{-1},\cdots\right\}$. Since $G$ is abelian the order of any product is irrelvant, therefore we may conclude that the product of all the elements of $G$ is $\underbrace{g\circ g^{-1}\cdot g_1\circ g_1^{-1}\cdots}_{n-1\text{ number of times}}e_G=\underbrace{e_G\circ e_G\cdots}_{\frac{n-1}{2}\text{ times}}e_G=e_G$

Remark: Obviously the parity of $|G|$ was important, for if it was even there would be no guarantee that some (or even all) elements of $G$ were there own inverse. In that case despite $G$ being abelian the product of all the unique elements of $G$ would not be $e_G$ for no element is repeated, which means that the element whose inverse is itself would not have an inverse in the product.
• Nov 12th 2009, 10:48 AM
signaldoc
"The group is finite. No element is its own invrse.

We know...but why?

Tonio"

Lagrange (as I previously posted): If a generates a subgroup of order 2, the order of the group would be even, contradiction...
• Nov 12th 2009, 11:10 AM
tonio
Quote:

Originally Posted by signaldoc
"The group is finite. No element is its own invrse.

We know...but why?

Tonio"

Lagrange (as I previously posted): If a generates a subgroup of order 2, the order of the group would be even, contradiction...

Oh...hehe, I know. I meant to ask the OP. Thanx anyway.(Wink)

Tonio
• Nov 13th 2009, 08:05 AM
biggybarks
help
Ok, I was told by my teacher that the way I did this is unacceptable. He wants me to show this without knowing how many elements are in the group. ex like 2k +1. He wants it more abstract. How do you know that none of the elements are their own inverse? In order to say that, for example, the inverse of a is one of the elements that comes later, you need to be sure a is not its own inverse. So you have to deal with that issue.

Any ideas now how to work this problem out?
• Nov 13th 2009, 09:23 AM
signaldoc
The complete argument is written out in the post by Drexel28 in this thread. It can be made more abstract notationally, I suppose, but the argument is good as it stands...

Maybe you can rewrite the product as follows. First prove that no element is its own inverse (see several posts already pointing this out). Then, given the product, P, of (all) the n elements g_i, rewrite it as the product, Q, of (n-1)/2 elements h_i, each of which is expressed as the product of two mutually inverse g_i. Stick the identity on Q so all elements are included. All you have done is rearrange a finite product. Now, since the group is abelian, P=Q. Moreover, each h_i is the identity. So Q=e. You may also note that in collecting pairs to form the h_i, each element g_i is used exactly once since inverses are unique.

Sorry I can't latex this for you...