# Thread: An example of abelian group where summation of a !=0 where a belongs to the group G

1. ## An example of abelian group where summation of a !=0 where a belongs to the group G

I am looking for an example of a finite abelian group (G) where the

Ʃa ǂ 0 where a ϵ G

Can someone help me with this one?

2. This really makes no sense:

$\sum a$

Can you explain the notation?

edit:

Did you mean:

$\displaystyle \sum_{a\in G} a \neq 0$?

3. Yes exactly I meant that...

4. Then how about $\mathbb{Z}/2\mathbb{Z}$?

5. I just cant imagine group theory scenarios, have been trying hard to think about the Z/2Z situation but cant understand why the summation under addition of all elements should not be the neutral element, the reason being the inverse of the element will also be present in the group right? so each element can be added to its inverse to give us the neutral element and in the end the summation of all the neutral element should give us the neutral element back...isn't it?

I am not getting much confidence in group theory, could you suggest how I can get better in it?

6. Another notation for $\mathbb{Z}/2\mathbb{Z}$ is $\mathbb{Z}_2 = \{0,1\}$. Are you sure you don't see how $\sum_{a \in \mathbb{Z}_2}{a}\ne 0$?

7. Originally Posted by xenonforlife
I just cant imagine group theory scenarios, have been trying hard to think about the Z/2Z situation but cant understand why the summation under addition of all elements should not be the neutral element, the reason being the inverse of the element will also be present in the group right? so each element can be added to its inverse to give us the neutral element and in the end the summation of all the neutral element should give us the neutral element back...isn't it?

I am not getting much confidence in group theory, could you suggest how I can get better in it?
Every element in a group has a unique inverse, yes. However, some elements are their own inverses (1 is its own inverse in the above group). Here is a more general approach to this problem, but certainly not the most general.

Take any even integer $n=2k$, and form the group $\mathbb{Z}/n\mathbb{Z}$. As SlipEternal mentioned, that quotient group can be thought of as the group $(\{0,1,...,n-1\}, +)$, where $+$ is addition mod n. Then what is the sum of the group elements?

$\displaystyle \sum_{i=0}^{n-1} i = \frac{(n-1)(n)}{2} = \frac{(n-1)(2k)}{2} = k(n-1)$.

Now $n-1$ is really the same element as $-1$, so in $G$ The sum is equal to $-k$. However $k$ is its own inverse, so the sum is just $k$ (which is not 0 if n>0).

For your last question, I was in your place once. I suggest you do every exercise you can. Many books give as exercises very central ideas. Confidence comes from experience.

Maybe think about exactly when it will happen that the sum is the identity, and do the proof.

8. Yep I kind of get it now, thank you both so so much for your inputs. , I believe Math can be best learnt as you said by practice and also by the exchange of ideas and different approaches to a problem. I am glad I bumped into this forum. Thanks again, I guess I will keep coming in as and when I am stuck, hoping that its not that often

9. ## Stuck again

Let G be a ﬁnite abelian group. If g ∈ G we write 2g for the element
g + g. Show that

2 Ʃa = 0 a ϵ G.

I am sorry I dont know how to use mathematical notations like the way you guys have mentioned in your replies. Please let me know how to do that too.

10. Originally Posted by xenonforlife
Let G be a ﬁnite abelian group. If g ∈ G we write 2g for the element
g + g. Show that

2 Ʃa = 0 a ϵ G.

I am sorry I dont know how to use mathematical notations like the way you guys have mentioned in your replies. Please let me know how to do that too.
This is very related to the more general example I have above. Recall that the example I gave had an element that was its own inverse. If an element g is its own inverse then g+g=0. It takes care of those "problem" elements who are their own inverses... Can you finish it now?

For the notation, double click on the image to see the code. Put that code inside: [tex.] [/tex.] without the periods.