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!