Consider the following statements: $𝑆_1$: If a group $(𝐺,∗)$ is of order $𝑛$, and $𝑎∈𝐺$ is such…

Question

Consider the following statements: $𝑆_1$: If a group $(𝐺,∗)$ is of order $𝑛$, and $𝑎∈𝐺$ is such that $𝑎^𝑚=𝑒$ for some integer $𝑚≤𝑛$, then $𝑚$ must divide $𝑛$. $𝑆_2$: If a group $(𝐺,∗)$ is of even order, then there must be an element $𝑎∈𝐺$ such that $𝑎≠𝑒$ and $𝑎∗𝑎=𝑒$. Which of the statements is (are) correct?

Accepted Answer

Correct answer: B — Answer: Only the statement asserting existence of a non-identity element whose square is the identity is correct. Explanation: The claim that if a^m = e for some m ≤ n then m must divide n is false. Counterexample: take a group of order 6 with an element of order 2. For that element a we have a^4 = e with m = 4 ≤ 6, but 4 does not divide 6. Reasoning: if d is the order of a, then d divides any exponent m for which a^m = e, and d divides |G| by Lagrange's theorem. However, a non-minimal exponent m need not divide |G|. The claim that any group of even order contains a non-identity element a with a*a = e is true. Proof idea: pair each element that is not its own inverse with its distinct inverse; these pairs contribute an even number of elements. The identity is its own inverse, so to make the total number of elements even there must be at least one other element that is its own inverse (an involution a ≠ e with a^2 = e).

Answer

A. Only $𝑆_1$

Answer

B. Only $𝑆_2$

Answer

C. Both $𝑆_1$ and $𝑆_2$

Answer

D. Neither $𝑆_1$ and $𝑆_2$

Consider the following statements: \(𝑆_1\): If a group \((𝐺,∗)\) is of order…