Let a and b be group elements such that |a|=2, b is not eual to e and aba=b^2. Find |b|.

Question

What I tried...

$$aba=b^2$$

$$a^2ba=ab^2$$

$$ba=ab^2$$ since $\mid a\mid=2$

$$ba^2=ab^2a$$

$$b=ab^2a$$ since $\mid a\mid =2.$

Stuck!

Answer 1

$$b^4=b^2b^2=abaaba=ab^2a=a^2ba^2=b$$

and thus $b^3=e$. Since $3$ is prime and $b\neq e$ then $|b|=3$.

Answer 2

A similar but different way of thinking about the solution is as follows. First square $b^2$. $b^4=ab(aa)ba=ab^2a$ Substituting $b^2$ in once again gives us $aabaa=b$. Thus $b^4=b.$ Acting on the equality using $b^{-1}$ gives $b^3=e$. We know $|b|\neq1$ since $b \neq e$. I will do a short proof by contradiction and subsequently conclude that $|b|=3$. Suppose that $b^2=e$. This implies that $aba=e$. Acting first on the left by $a$, and then on the right by $a$ gives $b=e$. But this contradicts the hypothesis. $\square$