If σ is a cycle of length n, then σ^r is also a cycle if and only if n and r are relatively prime If σ=(1.2.3), σσσ= identity permutation, which is cyclic

in this case n=3 and r=3 

but their gcd is not 1.

I don't understand why  -> this direction of theorem is true
By relabelling, we may suppose $\sigma = (1,2,\ldots, n)$.  Thus $\sigma(i) \equiv i+1 \mod n$.  Then $\sigma^r(i) \equiv i+r \mod n$.  This is not an $n$-cycle iff there is some $i$, $1 \le i \le n$, and some $k$, $1 \le k \le n-1$, such that $\sigma^{rk}(i) = i$.  That
says $i + r k \equiv i \mod n$, i.e. $r k \equiv 0 \mod n$.
If $g = \gcd(r,n) &gt; 1$, this will be true for $k = n/g$, but if $g = 1$, $r k \equiv 0 \mod n$ implies $k \equiv 0 \mod n$.
After working through the problem, I think you are right that $\gcd(n,r)=1$ for $\sigma^r$ to be a cycle. 

The example you provided isn't a counter example because the identity isn't a cycle. 

Instead, it is a product of cycles (each containing one element of $A$). So our condition holds true. 

Because $\gcd(n,r)=3$ in the example you described, we don't get a cycle, as the theorem predicts.