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It’s an exercise in my textbook.

Let $G$ be a finite Abelian group, then determine $$\prod\limits_{g\in G}g.$$

Actually, I do not quite get what it is asking. What does it mean by “determine”? What needs to be determined?

Moreover, there’s a following task asking me to show with the help of my first question that $$(p−1)! \equiv −1~{\rm mod}~p~~ (p~{\rm prime}).$$ I’m then getting more confused...

Any help, hint or detail, would be greatly appreciated. Thanks!

PS: They are exercises from my textbook, on page 49 of The Theory of Finite Groups, An Introduction (page 62 of the pdf).

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    https://math.stackexchange.com/questions/53185/the-product-of-all-the-elements-of-a-finite-abelian-group?rq=1 https://math.stackexchange.com/questions/9311/product-of-all-elements-in-an-odd-finite-abelian-group-is-1 and so on – Kiryl Pesotski Mar 26 '18 at 11:44
  • @KirylPesotski: Thanks! They look just great, I’ll read them carefully! –  Mar 26 '18 at 11:46
  • For what it's worth, determine in this context is hinting that there is a simpler or other way to understand what that product is. – Joppy Mar 26 '18 at 11:54
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    The way I see this, this question is not a duplicate (also see my answer). Actually I'm quite curious myself how this works for the general case case $k>3$. Might it be better to open a thread for this myself? – Václav Mordvinov Mar 26 '18 at 13:20
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    Perhaps if someone with at least 3000 rep agrees with us, they can vote to reopen – Václav Mordvinov Mar 26 '18 at 13:33
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    I decided to ask myself for the general case so an answer can be found here https://math.stackexchange.com/a/2709003/499176 – Václav Mordvinov Mar 26 '18 at 16:35
  • @VáclavMordvinov thanks a lot! –  Mar 26 '18 at 22:18

3 Answers3

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Every element $g \in G$ has an inverse $g^{-1}$. There are two cases:

$(i)$ $g=g^{-1}$, in which case $g$ is either the identity or an element of order $2$.

$(ii)$ $g \ne g^{-1}$.

In the product $\prod\limits_{g\in G}g$ each $g$ in case $(ii)$ can be paired with its inverse. This leaves the product of elements in case $(i)$.

When $G$ is $(\mathbb{Z}/p\mathbb{Z})^\times$ then

$\prod\limits_{g\in G}g = \prod\limits_{k=1}^{p-1}k \mod p=(p-1)! \mod p$

and each element in the product can be paired with its inverse apart from $1$ and $p-1$.

gandalf61
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  • Hello, thanks for your answer. Would you be so kind as to tell me how did you get $\prod g=\prod k$, where $g$ is an element but $k$ is a number? –  Mar 28 '18 at 12:07
  • The members of $(\mathbb{Z}/p\mathbb{Z})^\times$ are residue classes which can be represented by the integers from $1$ to $p-1$. – gandalf61 Mar 28 '18 at 12:59
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    Yes, I know that, I’m really sorry but I still cannot see how we could derive that equality. Would you be kind enough to give me more detail? –  Mar 28 '18 at 16:27
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    The integers from $1$ to $p-1$ are the elements of $G$. The group operation is multiplication modulo $p$. So the product of the groups elements is by definition the same as the product of the integers modulo $p$. We are just rewriting the same thing in a different way. – gandalf61 Mar 28 '18 at 20:22
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The answer of your first question has been pointed out by Kiryl Pesotski.

For your second question, notice that $(p-1)!$ is the product $\prod_{g \in G} g$ where $G$ is $(\mathbb{Z}/p\mathbb{Z})^\times$, and that $-1$ is of order $2$ in this group.

In this type of exercise, determine means reduce the product to an element you know ($e$ here) or to an element which has specific properties, that you must find.

Václav Mordvinov
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Dadedidododu
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Not a complete answer

A lot of your question is already answered by the links Kiryl Pesotski posted. The second question is already answered here and also in the links posted. There remains one case, however. For your first question:

  • If there is no element of order 2, every element and its inverse appear in the product, and the identity $e$ appears once, such that the product equals $e$. By Cauchy's theorem, this is the case when $\mathrm{order}(G)$ is odd.

  • If $\mathrm{order}(G)$ is even, again by Cauchy's theorem, there is an element of order $2$. Suppose there are $k$ elements of order $2$ and denote these by $g_1,g_2,\ldots,g_k$. Then, because $G$ is abelian by assumption, $\{e,g_1,g_2,\ldots,g_k\}\subset G$ is a subgroup (verify this). Then by Cauchy's theorem again, the order of this subgroup must be even, thus $k$ is odd. If we write out $\prod_{g\in G}g$ now, we observe that for all $g\not\in\{e,g_1,g_2,\ldots,g_k\}$, both the element and its inverse appear exactly once in the product, thereby yielding the identity element. Thus the product reduces to $\prod_{i=1}^{k}g_i$. For $k=1$, it's simple, $\prod_{g\in G}g=g_1$, for $k=3$ also: $\prod_{g\in G}g=g_1\circ g_2\circ g_3=g_3^2=e$, since $g_1\circ g_2\not\in \{e,g_1,g_2\}$. For any $k>3$ (the general case) the product reduces to exactly one element in $\{e,g_1,g_2,\ldots,g_k\}$ but to be honest I don't see how to infer anything about this.

So this is not a duplicate of the linked questions, the way I see it, because you are asked to determine the product in all generality. Maybe someone else knows how to calculate the product for $k>3$.

Václav Mordvinov
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