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I've come across this question and I tried to prove it, but my solution seems a little stealthy to me, is it correct?

Let $|G|=2k+1$, then we have:

$a=ae=aa^{|G|}=a^{|G|+1}=a^{2k+2}=(a^2)^{k+1}=(b^2)^{k+1}=b^{2k+2}=b^{|G|+1}=eb=b$

I would like to know if I'm missing something

Chain Markov
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ekosman
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2 Answers2

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Yep, that's a totally valid answer. Intuitively, the reason why this works is that in a group with an odd order, we can't have an element with an even order (by Lagrange's Theorem, trivially), so squaring something would never get us into a bind with the identity element that invalidates your $a = b \rightarrow a^2 = b^2$.

In fact, if you're interested in pursuing this a little further, you can think about why this approach could or couldn't prove that if $|G|$ has no common factors with some $n$, then $a^n = b^n \rightarrow a = b$. You can prove a more general result (though not necessarily one as clean as the one you have here!).

Robin Aldabanx
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Here is the generalization that Robin pointed out in his answer: If $(|G|, s) = 1$, then the map given by $$h \mapsto h^s$$ is a bijection. Although the argument presented by OP is quite slick, I couldn't apply it directly in this case. Here’s an alternative approach that works:

If $h_1^s = h_2^s$, then it follows that $|\langle h_1^s \rangle| = |\langle h_2^s \rangle| =: k$. Moreover, since $(|G|, s) = 1$, both subgroups $\langle h_1 \rangle$ and $\langle h_2\rangle$ have the same order $k$. Note that we immediately have from the observations above that $\langle h_1 \rangle = \langle h_2 \rangle$. Indeed, $\langle h_1 \rangle \cap \langle h_2 \rangle \subset \langle h_1 \rangle, \langle h_2 \rangle$ and contains an element of order $k$. Consequently, there exists some integer $t$ such that $0 < t < k$ and $h_1 = h_2^t$. Given that $h_1^{s} = h_2^{s}$ and their order is $k$, we have $$(h_2^{s})^{t} = h_1^{s} = h_2^{s}.$$ Thus, $t \equiv 1\operatorname{mod}\,k$, which yields $t = 1$.

There is also a converse. If $ (\lvert G \rvert, s) \neq 1 $, we can choose a prime $ p $ such that $p | |G|$ and $p | s$. By Cauchy's Theorem, there exists an element $ g \in G $ with order $ p $. Therefore, the function $h \mapsto h^{s}$ is not injective, as $g$ is mapped to $1$.

huh
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