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

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?

• Consider what happens to the identity element of your group. – Aeolian Feb 26 '13 at 1:34
• @Aeolian but the identity element has even order. – Ittay Weiss Feb 26 '13 at 1:37
• @IttayWeiss, the identity has order $1$. – Santiago Canez Feb 26 '13 at 2:14
• how so @SantiagoCanez ? – Ittay Weiss Feb 26 '13 at 4:31
• @IttayWeiss: The order of an element is by definition the smallest integer $n$ with $n \ge 1$ such that $g^n = 1$ (1 = identity element), or $\infty$ if there is no such $n$. So the order of the identity is 1. – Derek Holt Feb 26 '13 at 8:55

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 $k$.

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

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

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 $k$.

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

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 Hekster 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 $$|g|$$.