Let $G$ be a finite abelian group with exactly one element of order $2$ denoted by $\alpha$. Prove that $\prod_{g \in G} g = \alpha$.

Question

Let $G$ be a finite abelian group with exactly one element of order $2$ denoted by $\alpha$. Prove that $\prod_{g \in G} g = \alpha$

Okay so I was given two hints for this problem

Hint 1 : If $g \in G$ doesn't have order $2$, then $g \neq g^{-1}$

Hint 2: If you write out $0+1+2+3+4+50+1+2+3+4+5 \mod 6$, why does it come out $3$, other than computing brute force?

I'm not sure at all how to use hint 1.

The answer for Hint 2 is that you can pair up $5+1=6$ and $4+2 = 6$ and be left with $3$. In that hint we are looking at the additive group $\mathbb{Z}/6\mathbb{Z}$, and we have $|5| = |1| = 6$ and $|4| = |2| = 3$, so this hint would suggest looking at group operating each two elements of $G$ which have the same order with each other.

But the problem is, is that even though $G$ may be a finite group, some $g \in G$ may have $|g| = \infty$, and furthermore every $g \in G$ may have a unique order, (please correct me if I'm wrong). So I'm not sure how I could use it to prove $\prod_{g \in G} g = \alpha$

At this point all I can say about $G$, is that by Lagrange's Theorem, $|G| = 2k$ for some $k \in \mathbb{Z}^+$, and that (trivially) letting $\alpha \in G$ be the element of order $2$, we have $\alpha = \alpha^{-1}$.

I'm assuming there's some algebraic trick that I'm missing here, because I'm sure all I'd have to do is just write out $$\prod_{g \in G} g = g_1 \cdot g_2 \cdot \ ... \ \cdot g_{2k}$$ and find some way to simplify this expression.

EDIT: Thanks to the comments and answers below I realized that since $G$ is abelian, and for any $g \in G$, we have $g^{-1} \in G$, so $\prod_{g \in G} g = g_1 \cdot (g_1^{-1}) \cdot ... \cdot g_i \cdot(g_i^{-1}) \cdot ... \cdot (g_k) \cdot (g_k^{-1})$ where for some $i \in \{1, ..., k\}$ we have $g_i = \alpha$, but now I don't see why $\prod_{g \in G} g \neq e$ where $e$ is the identity of $G$?

Answer 1

Note that in your answer for hint 2, you are pairing $1$ with $5$ and $4$ with $2$. In the group $\mathbb{Z}/6\mathbb{Z}$, this exactly corresponds to pairing up inverse elements. Try to do this same pairing method for $G$. How many of the elements can be paired up in this fashion?

Answer 2

Alright, so as the problem says, lets let $G$ be a finite abelian group with one element, which we'll call $\alpha$, of order $2$.

Firstly, importatly recall the realization you made in the comments: $$\alpha^2=e \implies \alpha=\alpha^{-1}.$$

So now lets look at our product. We could go ahead and say $|G|$ is some postivie integer $n$ so we have something to refer to. Then $$\prod_{g\in G} g = g_1g_2g_3 \cdots \alpha \cdots g_{(n-1)}g_n$$ where each $g_i$ is a distinct element in $G$ and we know that $\alpha$, since it is also in $G$, is somewhere in that product.

As you realized thanks to the other answer, because this group is abelian we can shift things around. So lets go ahead and put every element next to it's inverse in the product -- but wait! This is where another thing we know comes into play: $\alpha$ is the only element of order $2$, that is, it's the only element whose inverse is itself.

Note that only elements of order $2$ have the property that their inverse is themself.

So now, rearranging our product, $\alpha$ is the only one doomed to live its life in solidarity without an inverse, because it's partner $\alpha^{-1}$ is nowhere to be seen, because it would require $\alpha$ to occur twice in the product: $$g_1g_1^{-1}g_2g_2^{-1}\cdots \alpha \cdots g_{\left(\frac{n-2}{2}-1\right)}g_{\left(\frac{n-2}{2}-1\right)}^{-1} g_{\frac{n-2}{2}}g_{\frac{n-2}{2}}^{-1}$$

(The reason the subscripts got wonky and went up to $\frac{n-2}{2}$ since two elements have the same subscript, but there's two odd one out, $e$ and $\alpha$. Don't worry too much about the subscripts here, just focus on everything having an inverse present except $\alpha$ (and $e$, but thats less important).)

So after canceling galore, we have:

$$\prod_{g\in G} g = ee \cdots \alpha \cdots ee = \alpha.$$