# Is it true that $A_x$ has even order for all $x\in G$?

Let $$G$$ be a finite group of odd order $$n$$.

Let $$x\in G$$.

Consider the set $$A_x=\{y\neq x:\gcd(o(x),o(y))=1 \text{or prime}\}$$.

Is it true that $$A_x$$ has even order for all $$x\in G$$?

Example: I considered the group $$\Bbb Z_5$$. Consider the element $$$$ .

Then $$A_{}=\{,,,\}$$

So it has even order.

NOTE:

From the comments of @Derek Holt I found that $$A_x$$ has even order if and only if $$o(x)$$ is $$1$$ or prime.

I am unable to sketch a proof of the above. Suppose that $$o(x)=1$$ or prime then it is easy to show that for any element $$y\in G$$, $$\gcd(o[x],o[y])=1$$ or prime and since $$G$$ has odd order and $$|A_x|=|G|-1$$ so $$A_x$$ is even.

How to prove the converse?

• So o(x) means order of element x? It is obvious not. For example that if n is prime and x the identify element, $A_x =G$ so that it has odd order n. – Zhaohui Du Sep 3 '19 at 11:17
• No, it has odd order for all $x \in G$, because it contains $y$ if and only if it contains $y^{-1}$. – Derek Holt Sep 3 '19 at 11:17
• @DerekHolt; If $y\neq x$ then is it true that $A_x$ has even order? – Math_Freak Sep 3 '19 at 15:28
• With your extra condition, $|A_x|$ is even if and only if $o(x) = 1$ or $o(x)$ is prime. – Derek Holt Sep 3 '19 at 16:21
• @Servaes; $o()=5$ and hence $\gcd(o,o)=5$ which is prime , so why do you say that $\notin A_{}$ – Math_Freak Sep 4 '19 at 2:13

## 2 Answers

I would give a proof of the converse of Derek Holt's conclusion.

If $$G$$ is of odd order then there is no element of even order. Therefore for $$A^*_x = \{y:gcd(o(x),o(y)) = 1 \text{or prime} \}$$ has properties for all $$x \in G$$ that

1. $$0 \in A^*_x$$

2. $$y \in A^*_x \leftrightarrow y^{-1} \in A^*_x$$. Remind that $$y \neq y^{-1}$$ except for $$y = 0$$ since $$o(y) \neq \text{even}$$.

Therefore $$A^*_x$$ is automatically of odd order for all $$x$$. If $$o(x) = 1 \text{or prime}$$ then $$x \in A^*_x$$ and $$A_x = A^*_x - {x}$$ then $$A_x$$ is of even order. Otherwise $$x \notin A^*_x$$ therefore $$A_x = A^*_x$$ is of odd order.

If $$G$$ is a finite abelian group then

$$H_{a,b} =\{x \in G, \ \gcd(ord(x),a)\ |\ b\ \}$$ is a subgroup.

(proof : $$ord(xy)\ |\ lcm(ord(x),ord(y))$$)

For $$p$$ prime $$\#\{x \in G, \gcd(ord(x),a) =p\}=|H_{a,p}|- |H_{a,1}|$$ Which is even if $$|G|$$ is odd since $$|H_{a,p}|, |H_{a,1}|$$ are odd.

Whence $$\#\{x \in G,\ \gcd(ord(x),a) \ is\ 1 \ or \ prime\ \} =|H_{a,1}|+ \sum_{p | a} (|H_{a,p}| -|H_{a,1}|)$$ is odd.