Question Let $G$ be a group of odd order. Then show that the diagonal elements of its multiplication table contain each element only once.

Attempt What I get ultimately is that $a^{2}c^{2} = e$ for some $a,c \in G$. And if my group were abelian, it would yield that $c$ is the inverse of $a$, but I don't see any contradiction here. Ideally, I feel I need to show if the diagonal could contain same elements twice, then somehow a subgroup of order two would exist, which would be a contradiction. Kindly help me go ahead.

  • 3
    $\begingroup$ Just a note this is part of a more general result: Let $G$ be order $n$, and $m$ an integer coprime to $n$. Then for every $g\in G$, the equation $x^m=g$ has a unique solution. Sketch of proof: show the map $x\mapsto x^m$ is surjective; since $G$ is finite, it is injective as well. $\endgroup$
    – Steve D
    May 14 at 23:56

2 Answers 2


Suppose $a^2 = b^2$. As $G$ has odd order there exists integers $s$ and $t$ such that $s \cdot |G| + t \cdot 2 = 1$.

Then $a = a^1 = a^{s \cdot |G|} \cdot (a^2)^t = (a^2)^t = (b^2)^t = b^{s \cdot |G|} \cdot (b^2)^t = b^1 = b$. Thus the map $g \rightarrow g^2$ is injective.

  • $\begingroup$ Thanks for the answer. very elegant.!! $\endgroup$
    – Debu
    May 15 at 5:34
  • 4
    $\begingroup$ Or $|G|=2n+1$ implies $a=a^{2n+2}=(a^2)^{n+1}=(b^2)^{n+1}=b^{2n+2}=b$ $\endgroup$
    – lhf
    May 15 at 10:31
  • $\begingroup$ Or even, $a=a^{2n+2}=(a^{n+1})^2$ proves that the map is surjective, hence also injective. $\endgroup$
    – lhf

Assume that you have elements $a$ and $b$ such that $a^2=b^2$.

Then $\langle a^2\rangle = \langle b^2\rangle$. But because $G$ has odd order, both $a$ and $b$ have odd order, so $\langle a^2\rangle = \langle a\rangle$ and $\langle b^2\rangle = \langle b\rangle$. So $\langle a\rangle = \langle b\rangle$. In particular, there exists $k$ such that $b=a^k$.

Then $a^2 = b^2 = (a^k)^2 = a^{2k}$. That means that $a^2=a^{2k}$. Therefore, $1=a^{2k-2} = a^{2(k-1)}$.

That means that the order of $a$ divides $2(k-1)$. Since the order is odd, it divides $k-1$. That means that $b=a^k = a^{(k-1)+1} = a^{k-1}a = a$.


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