Problem source

The elements $a,b,c,d$ belong to the group $G$ with binary operation
$*.$ Show that
\begin{questionparts}
\item if $a,b$ and $a*b$ are of order 2, then $a$ and $b$ commute; 
\item $c*d$ and $d*c$ have the same order; 
\item if $c^{-1}*b*c=b^{r},$ then $c^{-1}*b^{s}*c=b^{sr}$ and $c^{-n}*b^{s}*c^{n}=b^{sr^{n}}.$
\end{questionparts}

Solution source

\begin{questionparts}

\item $(ab)^2 = abab = e$ (since $ab$ has order $2$), but $a^2 = e, b^2 = e \Rightarrow a^{-1} = a, b^{-1} = b$ (since $a$ and $b$ have order 2) so $ba = ab$ by multiplication on the left by $a$ and right by $b$.

\item  Suppose $(cd)^n = e \Leftrightarrow d(cd)^nc = dc \Leftrightarrow (dc)^n(dc) = e \Leftrightarrow (dc)^n = e$

Therefore any number for which $(cd)^n = e$ has the property that $(dc)^n = e$ and vice-versa, in particular the smallest number for either $cd$ or $dc$ will also be the smallest number for the other.

\item Given $c^{-1}bc=b^r$, then $b^{rs} = (b^r)^s = (c^{-1}bc)^s =\underbrace{(c^{-1}bc)(c^{-1}bc) \cdots (c^{-1}bc)}_{s \text{ times}} = c^{-1}\underbrace{bb\cdots b}_{s \text{ times}}c = c^{-1}b^sc$

We proceed by induction on $n$. When $n = 0$, we have $b^s = b^{sr^0}$ so the base case is true.

Suppose it is true for some $n = k$, ie $c^{-k}b^sc^k = b^{sr^k}$. Now consider $c^{-{k+1}}b^sc^{k+1} = c^{-1}c^{-k}b^sc^kc = c^{-1}b^{sr^k}c = (b^{sr^k \cdot r}) = b^{sr^{k+1}}$ (where the second to last equality was by the previous part). Therefore if our statement is true for $n=k$ it is true for $n = k+1$. Therefore, since it is also true for $n=0$, by the principle of mathematical induction it is true for all non-negative integers $n$.