Exercise · ZW-SYLOW-THEOREMS-010 · ★★★ Advanced
A group of order \(12\) has \(n_3=4\). How many elements of order \(3\) does it have?
Choose one option.
Show answer
- B. \(8\)
Explanation
Each Sylow \(3\)-subgroup is cyclic of order \(3\) and contributes \(2\) elements of order \(3\); two distinct such subgroups meet only in the identity. That gives \(4\cdot2=8\) elements. The remaining \(4\) form the Sylow \(2\)-subgroup.
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-006 A group of order 21 has n_7=1. What follows?
- 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: