Question

(b) What conclusions can you draw from the result of part (a).

Short answer

The elements in the product term ${\mathrm{\alpha \beta \alpha }}^{-1}$depends on the nature of element$\mathrm{\beta }$, but the element of the product term${\mathrm{\alpha \beta \alpha }}^{-1}{\mathrm{\beta }}^{-1}$ is always even.

Step-by-step solution

Table from part (a) Theorem

$\mathrm{\alpha }$	$\mathrm{\beta }$	${\mathrm{\alpha \beta \alpha }}^{-1}$	${\mathrm{\alpha \beta \alpha }}^{-1}{\mathrm{\beta }}^{-1}$
even	even	even	even
even	odd	odd	even
odd	even	even	even
odd	odd	odd	even

Conclusion

From the table it can be observed that elements in the term${\mathrm{\alpha \beta \alpha }}^{-1}$ depends on the nature of element $\mathrm{\beta }$, but the element of the product term ${\mathrm{\alpha \beta \alpha }}^{-1}{\mathrm{\beta }}^{-1}$is always even.