Simple Property of GCD and Modular Arithmetic

Question

I'm stuck on proving a rather elementary property, as I'm not really sure how to start off the approach. Suppose $g^a\equiv 1$ mod $m$ and $g^b\equiv 1$ mod $m$. Does this imply that $g^{\gcd(a,b)}\equiv 1$ mod $m$?

Here's my attempt: By definition, we know that $m\mid g^a-1$ and $m|g^b-1$, so there exists some $x,y\in\mathbb{Z}$ such that $mx=g^a-1$ and $my=g^b-1$. Then \begin{align} m(x+y)&=g^{\gcd(a,b)}g^{\frac{a}{\gcd(a,b)}}+g^{\gcd(a,b)}g^{\frac{b}{\gcd(a,b)}}-2\\ &=g^{\gcd(a,b)}\Big(g^{\frac{a}{\gcd(a,b)}}+g^{\frac{b}{\gcd(a,b)}}\Big)-2. \end{align} However, I feel like this approach is only making the problem more complicated than it actually is, as the terms become harder and harder to manipulate to get our desired result $mz=g^{\gcd(a,b)}-1$ for some $z\in\mathbb{Z}.$

Any help would be appreciated!

Answer 1

We have by the extended Euclidean algorithm that there exists $x, y \in \Bbb{Z}$ such that $ax + by = gcd(a,b)$. So $$ g^{gcd(a,b)} = g^{ax+by} = g^{ax} \cdot g^{by} = (g^a)^x \cdot (g^b)^y \equiv 1^x\cdot 1^y = 1 \pmod{m}. $$

Answer 2

We do not even need Bézout's Theorem. All we need is to know that the original Euclidean Algorithm (with subtraction) terminates with the $\gcd$. Let $a\gt b$, and let $c=a-b$. Note that $\gcd(b,c)=\gcd(a,b)$.

We have $g^{c}g^{b}=g^{a}$, and therefore if $g^a\equiv 1\pmod{m}$ and $g^b\equiv 1 \pmod{m}$ then $g^c\equiv \pmod{m}$.

Answer 3

It is easy: $\rm\ \ g^A,\,g^B\equiv 1\:\Rightarrow\,order(g)\, |\, A,B\:\Rightarrow\: order(g)\, |\, (A,B)\:\Rightarrow\: g^{\,(A,B)}\equiv 1\quad$ QED