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As in the title. I found a related proof for groups of odd order here, but how can I extend it to groups of even order as well? There is also a very similar proof for groups where $3 \nmid |G|$ here.

Also, is it possible to avoid the fact that $a^{|G|} = e$?

Daniel Lai
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2 Answers2

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I don't believe it can be extended to even order because of the following counterexample:

Consider the cyclic group $G=C_4=\langle g\rangle$. Taking $a=g,b=g^3$ gives $a^2=g^2=b^2$, but $a\neq b$

Yaneda
  • 937
5

In general,

Proposition. Let $G$ be a finite group, $n$ a positive integer. The function (not a homomorphism) $f\colon G\to G$ given by $f(a)=a^n$ is one-to-one if and only if $\gcd(n,|G|)=1$.

Proof. If $\gcd(n,|G|)=1$, let $\alpha$ and $\beta$ be integers such that $\alpha n + \beta|G|=1$. Then for every $g\in G$ we have $$g = g^1 = g^{\alpha n+\beta|G|} = (g^{\alpha n})(g^{\beta|G|}) = (g^n)^{\alpha}(g^{|G|})^{\beta} = (g^n)^{\alpha}.$$ Therefore, if $a^n=b^n$, then $$a = (a^n)^{\alpha}= (b^n)^{\alpha} = b,$$ proving that $f$ is one-to-one.

Conversely, if $p$ is a prime that divides $\gcd(n,|G|)$, then by Cauchy's Theorem $G$ has an element $a$ of order $p$. Then $a^n=e$, since $p\mid n$, and therefore $a^n = e = e^n$. But $a\neq e$ since $a$ has order $p$, so $f(a)=f(e)$ but $a\neq e$, proving that $f$ is not one-to-one. $\Box$

Corollary. Let $G$ be a finite group, and let $n$ be a positive integer. Then following are equivalent:

  1. For all $a,b\in G$, $a^n=b^n$ implies $a=b$.
  2. $\gcd(n,|G|)=1$.

Corollary. If $|G|$ is odd and $a^2=b^2$, then $a=b$.

Corollary. If $3\nmid |G|$ and $a^3=b^3$, then $a=b$.

Now, you have the added condition of $a^n=b^n\neq e$. This introduces complications. For example, there are groups of even order in which $a^2=b^2\neq e$ implies $a=b$: in the Klein $4$-group, $x^2=e$ for all $x$, so the implication holds by vacuity. More generally, if every prime that divides $|G|$ also divides $n$, then there are groups $G$ in which $x^n=e$ for all $x\in G$ (even if $|G|$ itself does not divide $n$), so again the implication holds by vacuity.

But:

Theorem.

  1. Let $m$ and $n$ be positive integers such that $\gcd(n,m)\gt 1$ and assume there is a prime $p$ such that $p\mid m$ and $p\nmid n$. Then there exists a group $G$ such that $|G|=m$, and elements $a,b\in G$ such that $a\neq b$, but $a^n=b^n\neq e$.
  2. Let $m$ and $n$ be positive integers with $\gcd(n,m)\gt 1$. If for every prime $p$, $p\mid m$ implies $p\mid n$, then there exists a finite group $G$ of order $m$ such that for all $a,b\in G$, if $a^n=b^n\neq e$, then $a=b$.

Proof.

  1. Let $d=\gcd(n,m)$, and write $m=p^adm'$, with $\gcd(p,dm')=1$. Consider the abelian group $C_{p^a}\times C_d\times C_{m'}$; denote the generators of the cyclic factors by $x$, $y$, and $z$, respectively. Since $\gcd(p,n)=1$, then $x^n\neq e$. Since $d\mid n$, then $y^n=e$. Then, letting $a=x$ and $b=xy$, we have: $$a^n = x^n = x^ny^n = (xy)^n = b^n,$$ but since $y\neq e$, we have $a\neq b$.

  2. Write $m=p_1^{r_1}\cdots p_t^{r_t}$, with $p_i$ pairwise distinct primes, $r_i\gt 0$ for each $i$. Let $G$ be the abelian group $G= (C_{p_1})^{r_1}\times\cdots \times (C_{p_t})^{r_t}$. Then for every $g\in G$, we have $g^{p_1\cdots p_r}=e$. Therefore, for every $g\in G$ we have $g^n=e$. Thus, the implication $a^n=b^n\neq e\implies a=b$ hold by vacuity, since the antecedent never holds in $G$.~$\Box$

Arturo Magidin
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