Exercise · ZW-SYLOW-THEOREMS-006 · ★★☆ Intermediate
A group of order \(21\) has \(n_7=1\). What follows?
Choose one option.
Show answer
- C. its Sylow \(7\)-subgroup is normal
Explanation
With a unique Sylow \(7\)-subgroup, its conjugates coincide with it, so it is normal. It does not follow that the group is cyclic (a nonabelian group of order \(21\) exists) nor that the \(3\)-subgroup is normal, and a group with a proper normal subgroup is not simple.
Review the concept: Sylow theorems
A finite group \(G\) of order \(p^k m\), with \(p\) prime and \(p\nmid m\), has subgroups of order \(p^k\), all conjugate to each other; their number \(n_p\) is restricted.
\[n_p\equiv1\pmod p,\qquad n_p\mid m\]
More practice problems: sylow theorems
- ZW-SYLOW-THEOREMS-001 How many Sylow 5-subgroups can a group of order 15 have?
- ZW-SYLOW-THEOREMS-002 In a group of order 12, which is a possible value of n_3?
- ZW-SYLOW-THEOREMS-003 What is the order of a Sylow 2-subgroup of a group of order 72?
- ZW-SYLOW-THEOREMS-004 A group of order 56 has n_7 equal to:
- ZW-SYLOW-THEOREMS-005 A group G of order 20 has a unique Sylow 5-subgroup. How many elements of order 5 does G have?
- ZW-SYLOW-THEOREMS-007 How many Sylow 2-subgroups does the symmetric group S_3 have?
- ZW-SYLOW-THEOREMS-008 Every group of order 35 is:
- ZW-SYLOW-THEOREMS-009 If P is a Sylow 3-subgroup of G and g∈ G, the conjugate gPg^-1 is: