Discussion
The order of the smallest possible non trivial group containing elements \(x\) and \(y\) such that \(x^{7} = y^{2} = e\) and \(yx = x^{4}y\) is
Answer: B
The smallest non-trivial such group has order 2.
The relation \(yx = x^{4}y\) says \(yxy^{-1} = x^{4}\). Conjugating twice and using \(y^{2} = e\):
\[ \begin{aligned} x &= y^{2}xy^{-2} = (x^{4})^{4} = x^{16} \\ x^{15} &= e \end{aligned} \]
Together with \(x^{7} = e\), and \(\gcd(15, 7) = 1\), this forces \(x = e\). So the group is generated by \(y\) alone with \(y^{2} = e\), and the smallest non-trivial example is the cyclic group of order 2 (\(x = e\), \(y\) of order 2), which satisfies every relation.
Order 1 is the trivial group, which the question excludes. Orders 7 and 14 would need \(x\) of order 7, but \(x \mapsto x^{4}\) has order 3 as an automorphism of \(C_{7}\), so it cannot be conjugation by an element of order 2.
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