# Question regarding multiplication table of group of odd order.

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.

• 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. May 14 at 23:56

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.

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

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 $$b=a^k = a^{(k-1)+1} = a^{k-1}a = a$$.