Question about the symmetric group

Question

How do I prove that if $f\in S_k$ and $f^n=f^m=Id$, then $f^d=Id$ where $d=gcd(n,m)$?

I tried writing $f^{dn_1}=f^{dm_1}=f^0$ but this does not lead anywhere. I think I should use that $n_1$ and $m_1$ are prime to each other. I am not allowed to use that if $f^n=Id$ then the order of $f$ divides $n$. Any ideas?

Answer 1

Apply the Bezout identity: $$ \exists x,y\in\mathbb Z, xm+yn=d $$ $$ f^d=f^{xm+yn}=(f^m)^x(f^n)^y=id $$

Answer 2

We have $d=am+bn$ where $a$ and $b$ are integers. So, $f^d=f^{am}f^{bn}=\left(f^m\right)^a\left(f^n\right)^b=1$ because $f^n$ and $f^m$ are 1.