Let $P(a;b)$ be the set of all sequences generated by permutation of $a$ zeroes and $b$ ones. $$|P(a;b)|=\binom{a+b}{a}$$
I want to prove that if $gcd(a,b)=1$, every sequence in $P(a;b)$ will have $a+b$ unique circular shifts.
For the sake of brevity, I am defining an indexing set $\Lambda = \{ 0,1,\ldots,a+b-1 \}$ to order the elements of the sequences.
What is a circular shift? To apply circular shift on a sequence, remove the first element of the sequence and append it to the end. E.g. $a_1,a_2,\ldots, a_n$ becomes $a_2,a_3,\ldots, a_n, a_1$. Repeat this process iteratively. The sequences, hence obtained, are circular shifts of the original sequence.
Suppose we have a sequence, $\alpha=\left(\alpha_j : j\in \Lambda\right)$, its circular shifts are the sequences $\alpha^{(i)}=\left(\alpha_{i+j}: j\in \Lambda\right)$ taking subscripts modulo $a+b$.
After $a+b$ iterations of circular shifts, we get back $\alpha$ again i.e., $\alpha^{(a+b)}= \alpha$.
Applying circular shifts iteratively is similar to generating rotational symmetries of a regular polygon by rotating it at certain angle, (say, $90^\circ$ for squares).
Considering $\alpha$ as the correspondence to the identity element $e$ and applying circular shift $r$ iteratively, we obtain a group $\langle r\rangle$ where $r^{a+b}=e$.
From group theory, we know that order of $\langle r\rangle$ is either $a+b$ or some divisor of $a+b$.
Let $d$ be order of $\langle r\rangle$ where $a+b=d\times k$ for some $k\in \mathbb Z^+$.
Then, $r^d=e$ i.e., $\alpha^{(d)}=\alpha$. This is possible only if corresponding elements of these sequences are same.
$\alpha_0=\alpha_d=\alpha_{2d}=\ldots=\alpha_{(k-1)d}$
$\alpha_1=\alpha_{d+1}=\alpha_{2d+1}=\ldots=\alpha_{(k-1)d+1}$
and so on till
$\alpha_{d-1}=\alpha_{2d-1}=\alpha_{3d-1}=\ldots=\alpha_{kd-1}$.
We have partitioned the set $\Lambda$ into $d$ disjoint equivalence classes (each of size $k$) based on the relation $i\sim j$ iff $\alpha_i=\alpha_j$.
Any digit (be it zero or one) which occurs in $\alpha$ must do so in multiples of $k$. Thus, we need both $a$ and $b$ to be multiples of $k$. However, $gcd(a,b)=1$ so, the only possibility is $k=1$ and hence, $d=a+b$.
Thus, order of $\langle r\rangle$ is $d=a+b$ and hence, each of these $a+b$ circular shifts generated from $\alpha$ are unique. $\blacksquare$
Is my proof correct? Is there anything I need to clarify? I would love to see any alternate elementary proofs or improvement to this one.
I tried to look up this topic in this post, but couldn't understand it because I don't know anything about group actions yet.
