If G is a finite abelian group and $a_1,...,a_n$ are all its elements, show that $x=a_1a_2a_3...a_n$must satisfy $x^2=e$.

Question

I have already tried with $S_3$, and indeed, the product is $(13)$, and $(13)^2=e$

But what about this: I define + in this way:$45=2$,$26=3$, $1$ is the identity. therefore, $123456=2326=233=3$, and as you can see from the picture, the inverse of $3$ is not itself, thus contradicts the statement. My question:
What is wrong with my argument?
And How do you prove the statement?
it seems that my set of numbers perfectly forms a group under the operation I defined via the table above. first it's closed, second there is an identity, third every elements have an inverse. By the way it's commutitive. I tried (45)2=22=6=42=4(52), (45)6=26=3=43=4(56), so associativity seems to be satisfied.

Answer 1

$$a_1a_2a_3\ldots a_n a_1a_2a_3\ldots a_n$$ Let's start by canceling $a_1$. Some other, unique $a_i$ is equal to $a_1^{-1}$. Since the group is abelian, we can move that $a_i=a_1^{-1}$ from the second half over beside the $a_1$ in the first half (everything commutes, so order doesn't matter). So, those cancel, leaving $$a_2\ldots a_na_1\ldots a_{i-1}a_{i+1}\ldots a_n$$ Now we can do the same thing to cancel the $a_1$ on the right, we can just move it over next to the $a_i=a_1^{-1}$ on the left, and they cancel, so now we have $$a_2\ldots a_{i-1}a_{i+1}\ldots a_na_2\ldots a_{i-1}a_{i+1}\ldots a_n$$ Now both $a_1$ and $a_i$ are dealt with, and we're one step closer to the identity. We can do the same with $a_2,a_3,$ and so on.

What I'm writing isn't a rigorous proof, but do you get my drift as to why this works intuitively? Could you turn it into a rigorous proof?

Think about pairing each element with its inverse. Can you show that $x^2=a_1a_1^{-1}a_2a_2^{-1}\ldots a_na_n^{-1}a_1a_1^{-1}a_2a_2^{-1}\ldots a_na_n^{-1}$?

Extended question to consider: why is it that we can't just write $x=a_1a_2\ldots a_n=e$ using the above logic? What is the one thing that could mess that up and make us need another copy of $x$?

Answer 2

What you have defined is not a group.

Hint for your problem: Pair each $a_i$ that you can with its inverse, and multiply to get $1$. Which $a_i$ are left in your product?

Answer 3

Note that $x^2=(a_1a_2a_3\cdots a_n)(a_1a_2a_3\cdots a_n)=(a_1a_2a_3\cdots a_n)(a_1^{-1}a_2^{-1}a_3^{-1}\cdots a_n^{-1})$ because inversion is a bijection and the group is abelian.