Orders of $g^2$ when $g$ is odd

by Username Unknown   Last Updated May 23, 2020 01:20 AM

Suppose the order of $g$ is odd. What can you say about the order of $g^2$?

I think that the order of $g^2$ is going to be even. Is this true or false?

Tags : group-theory


Answers 6


Not necessarily. Consider G is ($\mathbb{Z}_5, +$) and $g=1$.

NECing
NECing
February 26, 2013 01:34 AM

Note: gcd(2, ord(g)) = 1. This might be germane.

ncmathsadist
ncmathsadist
February 26, 2013 01:34 AM

Suppose that $g$ has order $k$, where $k$ is odd. Then $g^k$ is the identity, and therefore $g^{k+1}=g$. Thus $(g^2)^{(k+1)/2}=g$.

That means that the subgroup generated by $g^2$ is the same as the subgroup generated by $g$. We conclude that $g^2$ also has order $k$.

André Nicolas
André Nicolas
February 26, 2013 01:39 AM

In general, if $ord(g)=n$, then $ord(g^k)=n/gcd(n,k)$. In particular, when $gcd(n,k)=1$, then $g$ and $g^k$ have the same order!

Nicky Hekster
Nicky Hekster
February 27, 2013 23:19 PM

Another way of showing the result that doesn't require going through showing the generated subgroups are the same: Let $k$ be the order of $g$ and $j$ be the order of $g^2$. Then we have $(g^2)^k = g^{2k} = (g^k)^2 = e^2 = e$, so $j\leq k$. To prove that it's not less than $k$, we break into two cases: $j\lt \frac k2$ and $\frac k2\lt j\lt k$ (note that $j=\frac k2$ is impossible since $k$ is odd - this is where that condition is used). If $j\lt \frac k2$, then $2j\lt k$; but we have $e=(g^2)^j=g^{2j}$, which contradicts the minimality of $k$ as the order of $g$. Similarly, if $\frac k2\lt j\lt k$, then $k\lt 2j \lt 2k$; then again $e=(g^2)^j=g^{2j} = g^{2j}\cdot g^{-k} = g^{2j-k}$ (where we use $g^{-k}=e$ in the middle equality). But the conditions imply that $0\lt 2j-k \lt k$, again contradicting the minimality of $k$.

Steven Stadnicki
Steven Stadnicki
February 27, 2013 23:44 PM

False, it will not be even. Furthermore, there's much more you can say about the order than whether it will even or odd in this case.

As some others have noted, if $g \in G$ is an element of finite order, then $$ |g^m| = \frac{|g|}{gcd(m, |g|)} $$

Since the order of $g$ is odd, $|g| = 2k + 1$ for some integer $k$. Furthermore, since we want to evaluate $|g^2|$, we have that $m = 2$. So,

$$ |g^2| = \frac{2k + 1}{gcd(2, 2k + 1)} $$

What's the $gcd$ of $2$ and $2k + 1$? It must be $1$ since odd numbers by definition are not divisible by $2$.

Hence,

$$ |g^2| = \frac{2k + 1}{1} = 2k + 1 = |g| $$

They have the same order! As Nicky Heckster notes in their more general answer https://math.stackexchange.com/a/316363/787867, anytime the $gcd$ of $m$ and $|g|$ is $1$, the order will be equivalent; that is, it will equal $|g|$.

Nicholas Montaño
Nicholas Montaño
May 23, 2020 00:42 AM

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