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Assume that $G$ is an abelian group of order $n$ and $a_1,a_2,...,a_n$ are its elements and we have $x=a_1a_2...a_n$.

If $G$ has more than one element such as $b\neq e$ and $b^2=e$, prove that $x=e$.

Answer:

Assume that $a\neq b$, $a^2=e$, $b^2=e$, $a\neq e$ and $b\neq e$. $$(ab)^2=(ab)(ab)=a^2b^2=e$$ If $$ab=e\implies a=b^{-1}\implies a=b$$ But we assume that $a\neq b$. contradiction!

So, $ab \neq e$ and $(ab)^2=e$.

So, what should I do now?

STUCK!

Shaun
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Sepehr Omidvar
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    That's Wilson's theorem for abelian groups https://en.wikipedia.org/wiki/Wilson%27s_theorem#Gauss's_generalization – lhf Dec 14 '20 at 16:17
  • Possible duplicate of https://math.stackexchange.com/questions/474214/product-of-elements-of-a-finite-abelian-group and others – lhf Dec 14 '20 at 16:23
  • $ab$ is also an element of $G$ and is different than $a$ and $b$, so it must appear in the product, now use this hint to solve the problem – Laassila souhayl Dec 14 '20 at 16:24
  • Inverses come in pairs (except when $a^{2} = e$). Because your group is abelian you can all cancel them out. Now look at what is left over. – Elmex80s Dec 14 '20 at 16:55
  • If you are stuck, then often this site has really good solutions for your problem. Take a look at the answers of older posts, too. – Dietrich Burde Dec 14 '20 at 19:18

2 Answers2

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$\newcommand{\Set}[1]{\left\{ #1 \right\}}$The main point is what happens in the subgroup $$ H = \Set { x \in G : x^{2} = e }, $$ which by assumption is a vector space of dimension $k$ over the field $F$ with $2$ elements.

Write the elements of $H$ with respect to a (multiplicative) basis. Looking at the exponents with respect to any given basis elements $b$, half of the elements of $H$ will have exponent $0$, and half will have exponent $1$. But half of $2^{k}$ is $2^{k-1}$, which is even as by assumption $k > 1$, so the sum $s$ of the exponents will be even, so that $b^{s} = e$.

Andreas Caranti
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Note that the elements $x \in G$ such that $x^2 = e$ are a subgroup, call it $H$.

Every element in this group has order $2$. By the classification of finite abelian groups, we must have $H \cong (\mathbb{Z}/2 \mathbb{Z})^k$ for some $k$. The statement is clearly true for $H$ (the product of all elements is the identity).

Now consider the elements in $G \setminus H$. We can pair each element with their inverse to get product $e$, since no element is its own inverse.

Putting these two pieces together, you get that the product of all the elements is the identity.

Sofía Marlasca Aparicio
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