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Let $G = \{a_1, \cdots , a_k \}$ be a finite abelian group. Show that $a_1a_2 \cdots a_k$ has order $1$ or $2$.

So far I have done this:

I think that the product $a_1a_2 \cdots a_k$ will have order $1$ when none of the elements of $G$ is inverse of itself. since by reorganizing the elements in an adequate way we arrive at the identity. Otherwise, if for some $a_i$ in $G$ $a_i$ is inverse of itself, then the product will have order $2$. If so, how could you formalize this?

Hopmaths
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2 Answers2

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Let $I$ be the subset of $G$ such that $a_i\in I$ implies that $a_i\neq a_i^{-1}$ and $J$ the subsets of $G$ such that $b_i\in J$ implies that $b_i=b_i^{-1}$. The product of elements of $G$ is the product $b_1...b_l$ of elements of $J$, we have $(b_1..b_l)^2=(b_1)^2..(b_l)^2=1$.

Tsemo Aristide
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  • Did I say something meaningful to my question? (To be able to guide me) – Hopmaths Dec 22 '20 at 01:33
  • @Hopmaths this hint more or less nails it, the last item remaining is figuring out how it can have order $2$ :) – gt6989b Dec 22 '20 at 01:37
  • Indeed the elements of order dividing $2$ form an $\Bbb F_2$-vector space, and we can use this to show $\prod g$ has order $2$ if and only if $G$ has precisely one element of order $2$. – anon Dec 22 '20 at 05:47
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For any finite abelian group $G=:\{g_1,\dots,g_k\}$, there are $n\in \Bbb N$ and positive integers $l_1,\dots,l_n$ such that $G\cong \Bbb Z_{l_1}\times\dots\times \Bbb Z_{l_n}=:\{a_1,\dots,a_k\}$, with $k=\prod_{i=1}^nl_i$ (fundamental theorem of finite abelian groups). In additive notation, we get:

\begin{alignat}{1} \sum_{i=1}^{k}a_i &\stackrel{\dagger}{=} \biggl(l_2\dots l_n\sum_{i=0}^{l_1-1}i,\space\space\dots,\space\space l_1\dots l_{n-1}\sum_{i=0}^{l_n-1}i\biggr) \\ &= \biggl(l_2\dots l_n\frac{(l_1-1)l_1}{2},\space\space\dots,\space\space l_1\dots l_{n-1}\frac{(l_n-1)l_n}{2}\biggr) \\ \tag 1 \end{alignat}

whence:

\begin{alignat}{1} 2\sum_{i=1}^{k}a_i &= \bigl(l_2\dots l_n(l_1-1)\color{red}{l_1},\space\space\dots,\space\space l_1\dots l_{n-1}(l_n-1)\color{red}{l_n}\bigr) \\ &= (0,\dots,0) \tag 2 \end{alignat}

Therefore, the element $\sum_{i=1}^{k}a_i$ has order at most $2$, and so does its "isomorphic image" $\sum_{i=1}^{k}g_i$.

$\dagger$ See here.