dw16-7 This is Daily Work 16. Current Students: Enter your BSU email address to receive a copy of your responses. Let \(G\) be a mysterious finite group of order 144, and suppose that \(H\) is a subgroup of \(G\).Which of the following cannot be the order of \(H\)? 1. 24. 108. 144. The symmetric group \(S_4\) has order 24. Does it have a subgroup of order 12? Yes, because 12 divides 24 we can apply Lagrange's theorem. Yes, because there is always a subgroup having half the elements of a group. No, because the converse of Lagrange's theorem is false. Lagrange's theorem cannot tell us either way. Which line of this proof contains an error?Let \(G\) be a group of order 9. We'll show \(G\) is cyclic.I. Let \(a\in G\) be an arbitrary element.II. By Lagrange's theorem, \( |a| = 9 \).III. Thus \( \langle a \rangle \) is a subgroup of order 9.IV. So \( \langle a \rangle = G \) and \(G\) is cyclic. Line I. Line II. Line III. Line IV. Related