Discussion
Let \(G = \{e, x, x^{2}, x^{3}, y, xy, x^{2}y, x^{3}y\}\) with \(o(x) = 4,\ o(y) = 2\) and \(xy = yx^{3}\). Then the number of elements in the center of the group \(G\) is equal to
Answer: B
The centre has 2 elements.
The relations \(o(x) = 4\), \(o(y) = 2\) and \(xy = yx^{3} = yx^{-1}\) describe the dihedral group of order 8, the symmetries of a square, with \(x\) a rotation and \(y\) a reflection.
An element \(x^{k}\) commutes with \(y\) only if \(x^{k} = x^{-k}\), that is \(k = 0\) or 2. No element \(x^{k}y\) commutes with \(x\), since \(x \cdot x^{k}y = x^{k+1}y\) while \(x^{k}y \cdot x = x^{k-1}y\). So \(Z(G) = \{e, x^{2}\}\).
8 would need G to be abelian, which it is not; a centre of order 4 would make \(G/Z(G)\) cyclic of order 2, forcing G to be abelian; and a non-trivial 2-group cannot have centre of order 1.
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