Exercise · ZW-SYLOW-THEOREMS-008 · ★★★ Advanced
Every group of order \(35\) is:
Choose one option.
Show answer
- A. cyclic
Explanation
Here \(n_5\) divides \(7\) and is congruent to \(1\) modulo \(5\), so \(n_5=1\); likewise \(n_7=1\). Both subgroups are normal, intersect trivially and their product is all of \(G\), so \(G\cong\mathbb{Z}_5\times\mathbb{Z}_7\cong\mathbb{Z}_{35}\).
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-009 If P is a Sylow 3-subgroup of G and g∈ G, the conjugate gPg^-1 is: