# abelian groups: proving $\prod\limits_{g\in G}g=\prod\limits_{g\in G}_{g^2=1}g$

Let $G$ be a finite abelian group. Why is $$\prod\limits_{g\in G}g=\prod\limits_{g\in G}_{g^2=1}g$$? I've tried to consider an element with its inverse but I don't get why you can combine the elements - which are different of its inverse - and they annul themselves in the product. Any help?

• I just thought that a comment was in order to mention that in fact the product mentioned is $1$ unless the group has a unique element of order $2$, in which case, it of course becomes that element. – Tobias Kildetoft Oct 19 '12 at 20:15

For any two different elements $a$, $b$ such that $ab=1$, pair off $a$ and $b$. So you have a bunch of couples, and some loners. The product of all elements that have been coupled off is $1$. What's left is the people who are their own inverses. Note that $g$ is its own inverse iff $g^2=1$.

We have $$\prod_{g\in G}g = \prod_{\substack{g\in G\\g^2=1}}g \cdot\prod_{\substack{g\in G\\g^2\neq1}}g$$ In the product $\displaystyle\prod_{\substack{g\in G\\g^2\neq1}}g$ every element cancels with its inverse.

$\prod\limits_{g\in G}g=\prod\limits_{g\in G}_{g^2=1}g = g_1 g_2 \dots g_n$

For some elements in the product $\prod\limits_{g_i \in G}g_i$ we have $g_i g_j = 1$. Since the group is Abelian you can arrange the product so that $g_i$ and $g_j$ are "next to each other" in the product. Then

$g_1 g_2 \dots g_n = g_{i_1} g_{j_1} g_{i_2}g_{j_2} \dots g_k \dots g_{k+k'} \stackrel{\ast}{=} g_k g_{k+1} \dots g_{k +k'} = \prod\limits_{g\in G}_{g^2=1}g$.

where in $\stackrel{\ast}{=}$ we use that $g_{i_1} g_{j_1} = 1 = g_{i_l}g_{j_l}$ for all elements where $g_i \neq g_i^{-1}$.

Here is a formal proof (with the same content as the other answers). The method is useful in other contexts, too.

Define the relation $\sim$ on $G$ by $a \sim b \Leftrightarrow a = b \vee a = b^{-1}$. One checks that this is an equivalence relation. Hence $G$ is the disjoint union of the equivalence classes. Since $G$ is abelian, it follows

$\prod_{a \in G} a = \prod_{[a] \in G/\sim} \prod_{b \in [a]} b$

We have $[a]=\{a,a^{-1}\}$, which has either two elements or just one element when $a=a^{-1}$ resp. $a^2=1$. In the first case, $\prod_{b \in [a]} b = a \cdot a^{-1} = 1$. In the second case, the product is $a$. Thus the whole product reduces to $\prod_{a \in G, a^2=1} a$.

Although this is offtopic, let me mention an application of this equivalence relation: If $G$ is a finite group whose order is even, then $G$ contains an element of order $2$ (and a similar proof works for arbitrary primes, this is Cauchy's Theorem). Namely, the class $[1]$ has just one element, so that at least one other class must have only one element.

It's exactly what you have started: if $g\ne g^{-1}$ then they will vanish from the left side.

• because $g\cdot g^{-1}=1$. – Berci Oct 19 '12 at 18:43