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Let $G=\{g_1,...,g_n\},n\in \mathbb{N}$ be an Abelian group and $c=g_1...g_n$. Show:

  1. $c^2=e$.
  2. if $n$ is odd, then $c=e$.
  3. if there exists unique $g\in G$ such that $ord(g)=2$, then $c \neq e$.

I tried multiple things for 1. for example, but nothing worked: I look at $c^2=g_1...g_ng_1...g_n=g_1...g_n(g_n^{-1}...g_1^{-1})^{-1}$, but here I was lost. Then I started to think that because $G$ is group $c=g_1...g_n\in G$ and so there exists $i\in \{1,...,n\}$, such that $g_i=g_1...g_n$ or $g_i=g_1...g_{i-1}g_{i+1}...g_ng_i$ and from that I got $g_1...g_{i-1}g_{i+1}...g_n=e$, but here I was lost once again. Any idea on how to start?

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    Hint: if $G={g_1,\dots,g_n}$, then for every $g_i$ must exist one inverse in $G$. Now, if you are multiplying all elements of $G$, you are considering elements and their inverse. Can you see that you can rearrange the product in a convenient way because G is abelian? And you have to consider the elements $g$ such that $g^2=e$. Now can you see why you have to square and why if n is odd then $c=e$? The third is obvious. – gamer rossi Nov 19 '24 at 23:18
  • I now see why 1. and 2. hold... But why does 3. hold, can you explain it a bit? –  Nov 19 '24 at 23:31
  • If there's a unique $g$ with order $2$, then pair up each of the other elements with its inverse. – Robert Shore Nov 19 '24 at 23:50
  • I don't understand that last part. I tried somehow by contradiction but can't get through –  Nov 20 '24 at 00:04
  • I didn't really done part 2 I figure it out. Can someone explain 2. and 3. in details? –  Nov 20 '24 at 00:16
  • For 3, note that $\prod_{i=1}^ng_i=\prod_{i=1,o(g_i)=2}^ng_i$. Use that $G$ is abelian and match each element of order different from 2 (this is, elements which are not its own inverse) with its inverse. – Deif Nov 20 '24 at 00:32
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    See also Wilson's theorem, and Gauss's generalization. – suckling pig Nov 20 '24 at 02:17
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    Duplicate question. Use site search with terms like product elements abelian group. The question been asked multiple times over the years. – Martin Brandenburg Nov 20 '24 at 02:33
  • This needs focus and a few of the questions have definitely been answered here before. – Shaun Nov 20 '24 at 13:11

2 Answers2

1

  1. We can pair up each element with its inverse in the product, which reduces it to the identity.

  2. This can be solved using Lagrange's Theorem, which tells us that in a group of odd size, there cannot be any element of order $2$. Note that $g^2=e\iff g=g^{-1}$, so for a non-identity element, having order 2 is equivalent to the element being equal to its inverse. Hence, all elements can be paired up with a distinct inverse in $c=g_1g_2\cdots g_n$, reducing it to $e$. If you want a proof without Lagrange's theorem, see here.

  3. Since $g$ is the unique element of order two, $g^2=e\iff g=g^{-1}$ once again implies that all other non-identity elements have each have an inverse that's not itself, so they can be paired up in $c=g_1g_2\cdots g_n$. Thus, they cancel out, leaving behind just $g$, which is not equal to $e$.

masky
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In addition to the comments above, there is also a very neat answer to the general question $\color{blue} {\text{What is the set of all different products of all the elements of a finite group $G$? }}$ So $G$ not necessarily abelian and in a product each element appears exactly once.

Well, if a $2$-Sylow subgroup of $G$ is trivial or non-cyclic, then this set equals the commutator subgroup $G'$.

If a $2$-Sylow subgroup of $G$ is (non-trivial) cyclic, then this set is the coset $xG'$ of the commutator subgroup, with $x$ the unique involution of a $2$-Sylow subgroup.

See also J. Dénes and P. Hermann, On the product of all elements in a finite group, Ann. Discrete Math. 15 (1982) 105-109. The theorem connects to the theory of Latin Squares and so-called complete maps.

Nicky Hekster
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