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Chapter 9: Q9.3-6E-c (page 303)

Classify all groups of the given order:391

Short Answer

Expert verified

The group of order 391 is $ℤ_{391}$ .

Step by step solution

01

Step-by-Step Solution Step 1: Referring to Corollary 9.18

Let G be a group of order pq, where p and q are primes such that p>q.

If $q \times (p - 1)$ , then $G ≅ ℤ_{p q}$ .

02

Classifying group of order 391

Writing the 391 as multiple of prime factors as:

$391 = 17 \cdot 23$

Therefore, by referring tocorollary 9.18, we get p=23 and q=17 (both are primes).

Hence, the possible group of the given order is ${ℤ_{p}}_{q} = ℤ_{391}$ .

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Most popular questions from this chapter

Show that T is a nonabelian subgroup of G.

If N is a subgroup of $Z (G)$ , prove that N is a normal subgroup of G .

Let $N_{1}, \dots, N_{k}$ be subgroups of an abelian group G.Assume that every element of G can be written in the form role="math" localid="1653628920687" $a_{1} \dots a_{n}$ (with $a_{i} \in N_{i}$ ) and that whenever role="math" localid="1653628977564" $a_{1} a_{2} \dots a_{n} = e$ , then $a_{i} = e$ for every i . Prove that $G = N_{1} \times N_{2} \times \dots N_{k}$ .

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Let $G$ be a group and $f_{1} : G \to G_{1}, f_{2} : G \to G_{2}, \dots, f_{n} : G \to G_{n}$ homomorphisms. For $i = 1, 2, \dots, n$ , let $π_{i}$ be the homomorphism of Exercise 8. Let $f^{*} : G \to G_{1} \times \dots \times G_{n}$ be the map defined by $f^{*} (a) = (f_{1} (a_{1}), f_{2} (a_{2}), \dots, f_{n} (a_{n}))$ .

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