Question

Suppose that \( G \) is a group of order 57 which is not cyclic. If \( G \) contains a unique subgroup \( H \) of order 19, then for any \( g \notin H \), the order of \( g \) is ................

Hint:

If \( H \) is a unique normal subgroup, elements outside it correspond to cosets forming a quotient group whose order equals the index of \( H \).

Answer 1

Correct Answer: 3

Step 1: Factorize the group order.
\[ |G| = 57 = 3 \times 19. \]

Step 2: Use Sylow's theorems.
If \( G \) has a unique subgroup \( H \) of order 19, \( H \) is normal.

Step 3: Consider quotient group \( G/H \).
\[ |G/H| = \frac{|G|}{|H|} = \frac{57}{19} = 3. \] Thus, \( G/H \) is cyclic of order 3.

Step 4: Order of element outside \( H \).
Any \( g \notin H \) corresponds to a non-identity element of \( G/H \), so its order in \( G/H \) is 3. Hence, \( o(g) = 3. \)

Final Answer: \[ \boxed{3} \]