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Could someone tell me if this proof I made is correct?

Let $(G,\cdot)$ be a finite abelian group. Show that $\prod_{g\in G}g^2=1$.

Proof

Suppose $G$ is a finite abelian group of order $n$, and let $G=\{1,g_1,...,g_k,g_{k+1},...,g_{n-1}\}$, $k< n$, where $1$ is the identity.
Suppose $1^2=1$, $g_i^2=1$ for $1\leq i\leq k$, $g_{k+1}=(g_{n-1})^{-1}$, $g_{j+1}=(g_j)^{-1}$ for $k+1\leq j \leq n-2$.
Then: $$\begin{equation}\begin{aligned} \prod_{g\in G}g^2 &= 1^2\cdot g_1^2\cdot\ ...\ \cdot g_k^2 \cdot g_{k+1}^2\cdot \ ... \ \cdot g_{n-1}^2\\ &=1^2\cdot g_1^2\cdot\ ...\ \cdot g_k^2 \cdot g_{k+1}\cdot g_{k+1}\cdot \ ... \ \cdot g_{n-1}\cdot g_{n-1}\\ &= 1^2\cdot g_1^2\cdot\ ...\ \cdot g_k^2\cdot (g_{n-1}\cdot g_{k+1})\cdot (g_{k+1}\cdot g_{k+2})\cdot \ ... \ \cdot (g_{n-2}\cdot g_{n-1})\\ &= 1^2\cdot g_1^2\cdot\ ...\ \cdot g_k^2\cdot (g_{n-1}\cdot (g_{n-1})^{-1})\cdot (g_{k+1}\cdot (g_{k+1})^{-1})\cdot \ ... \ \cdot (g_{n-2}\cdot (g_{n-2})^{-1})\\ &= 1\cdot 1\cdot \ ...\ \cdot1\cdot1\cdot1\cdot \ ... \ \cdot1\\ &= 1. \end{aligned}\end{equation}$$

Martin Brandenburg
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LM5
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    For future questions: This site is not a proof checking machine. Please check out the [tag:solution-verification] tag description: For posts looking for feedback or verification of a proposed solution. "Is this proof correct?" is too broad or missing context. Instead, the question must identify precisely which step in the proof is in doubt, and why so. This should not be the only tag for a question, and should not be used to circumvent site policies regarding duplication. And in fact, this problem is a (x-times) duplicate. – Martin Brandenburg Jan 30 '25 at 23:22

1 Answers1

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I believe that your proof is incorrect. You assume that $g_{k+1}=g_{n-1}^{-1}$ and $g_{k+2}=g_{k+1}^{-1}$ which would imply that $g_{k+2}=g_{n-1}$, which does not seem to be what you intended.

Instead the proposition can be proven as follows:

LEMMA let $(G,\cdot)$ and $(H,\cdot)$ be abelian groups satisfying $\prod_{g\in G}g^2=1_G$ and $\prod_{h\in H}h^2=1_H$ then $\prod_{q\in G\times H}q^2=1_{G\times H}$.

proof: rewrite $q\in G\times H$ as $q=(g,h)$ with $g\in G$ and $h\in H$. Then the product can be rewritten as follows \begin{align} \prod_{q\in G\times H}q^2&=\prod_{g\in G}\prod_{h\in H}(g,h)^2\\ &=\prod_{g\in G}\prod_{h\in H}(g^2,h^2)\\ &=\prod_{g\in G}\left(g^2,\prod_{h\in H}h^2\right)\\ &=\prod_{g\in G}(g^2,1_H)\\ &=\left(\prod_{g\in G}g^2,1_H\right)\\ &=(1_G,1_H)\\ &=1_{G\times H} \end{align} $\blacksquare$

The main proposition can then be proven using the structure theorem for finite abelian groups.

proposition Let $(G,\cdot)$ be a finite abelian group. Then $\prod_{g\in G}g^2=1$.

proof: since $G$ is a finite abelian group it follows from the structure theorem for finite abelian groups that, for some $n\in \mathbb N$, there are prime numbers $p_1,p_2,\ldots,p_n$ and natural numbers $k_1,k_2,\ldots,k_n$ such that $$G\cong \mathbb Z_{p_1^{k_1}}\times\mathbb Z_{p_2^{k_2}}\times\cdots\times\mathbb Z_{p_n^{k_n}}.$$ Define $\varphi_1,\varphi_2,\ldots,\varphi_n$ by $\varphi_i:\mathbb Z\to\mathbb Z_{p_i^{k_i}}$ where $\varphi_i:m\mapsto [m]_{p_i^{k_i}}$ for all $1\leq i\leq n$, notice also that all the $\varphi_i$'s are group homomorphisms.

It is then seen that, for all $1\leq i\leq n$, it holds that \begin{align} \varphi_i\left(\sum_{j=1}^{p_i^{k_i}-1}2j\right)&=\varphi_i \left(2\sum_{j=1}^{p_i^{k_i}-1}j\right)\\ &=\varphi_i\left(2\cdot\frac{p_i^{k_i}(p_i^{k_i}-1)}{2}\right)\\ &=\varphi_i(p_i^{k_i}(p_i^{k_i}-1))\\ &=[p_i^{k_i}(p_i^{k_i}-1)]_{p_i^{k_i}}\\ &=[0]_{p_i^{k_i}}. \end{align}

It is then seen that all the $\mathbb Z_{p_1^{k_1}},\mathbb Z_{p_2^{k_2}},\ldots,\mathbb Z_{p_n^{k_n}}$ satisfy the product property described in the above lemma, and since $G$ is isomorphic to their product it the follows by the lemma that $G$ also obeys the product property. $\blacksquare$

RBP
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