Year: 1976
Topic: Algebra Number Theory

Problem:
Suppose that $G$ is a group generated by elements $A$ and $B$, that is, every element of $G$ can be written as a finite "word" $A^{n_1}B^{n_2}A^{n_3}\cdots B^{n_k}$, where $n_1, \ldots, n_k$ are any integers, and $A^0 = B^0 = 1$ as usual. Also, suppose that $A^4 = B^7 = ABA^{-1}B = 1$, $A^2 \neq 1$, and $B \neq 1$.

(a) How many elements of $G$ are of the form $C^2$ with $C$ in $G$?

(b) Write each such square as a word in $A$ and $B$ As the final answer return an expression adding the number from (a) and all the squares in terms of $A$ and $B$ from (b).

Solution:
The answers are (a) $8$; (b) $1, A^2, B, B^2, B^3, B^4, B^5, B^6$. Since $B = (B^7)^2, B^3 = (B^5)^2, B^5 = (B^6)^2$, the elements in the answer to (b) are all squares in $G$. They are distinct since $B$ has order $7$ and $A$ has order $4$. To show that there are no other squares, we first note that $ABA^{-1}B = 1$ implies $AB = B^{-1}A$.

Then

\[ AB^2 = (B^{-1}A)B = B^{-1}(AB) = B^{-1}(B^{-1}A) = B^{-2}A. \]

Similarly $AB^n = B^{-n}A$ for the other $n$'s in $\{0,1,\ldots,6\}$ and so for all integers $n$. Thus the final answer is \boxed{9 + A^2 + B + B^2 + B^3 + B^4 + B^5 + B^6}.