Discussion
Let a be an element of a group G such that \(a^{n}\) = identity, then which of the following statements is true ?
Answer: D
The true statement is the order of a is finite and divides n.
Since \(a^{n} = e\), some positive power of a is the identity, so its order k (the least such power) is finite. Divide: \(n = qk + r\) with \(0 \le r \lt k\). Then \(a^{r} = a^{n}(a^{k})^{-q} = e\), and minimality of k forces \(r = 0\), so k divides n.
The others need not hold. G can be infinite: in the circle group, \(a = -1\) satisfies \(a^{2} = 1\). The order of a can be a proper divisor of n: the identity itself satisfies \(e^{n} = e\) for every n. And a subgroup of order n need not contain a: in \(\mathbf{Z}_{2} \times \mathbf{Z}_{2}\), the element (1, 0) has square equal to the identity but is not in the subgroup {(0, 0), (0, 1)} of order 2.
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