# Understand the problem

The multiplicative group \(F^*_7\) is isomorphic to a subgroup of the multiplicative group \(F^*_{31}\).

##### Source of the problem

TIFR GS 2018 Part A Problem 17

##### Topic

Abstract Algebra

##### Difficulty Level

Medium

##### Suggested Book

Dummit and Foote

# Start with hints

Do you really need a hint? Try it first!

We will write them as (Z/7Z)* and (Z/31Z)* respectively instead of the notations used.

- Observe that (Z/7Z)* has order 6 and (Z/31Z)* has order 30.So there is a possibility that (Z/7Z)* is a subgroup of (Z/31Z)* by Lagrange’s Theorem.
- So we need to go into the structure of the groups to solve this problem.Hence we proceed!
- Let us investigate the group (Z/7Z)*.It consists of {1,2,3,4,5,6 mod 7}.Observe that 3 mod 7 generates the group.
- So naturally the next question is that whether (Z/31Z)* rather is there any general result?

- In fact the following theorem is true and describes the cyclicity (Z/nZ)* to some extent.
- Theorem: If p is a prime then (Z/pZ)* is cyclic. (Check!) {Check the bonus question for the complete characterization of cyclicity of (Z/nZ)* done by Gauss.}

- So (Z/7Z)* and (Z/31Z)* are cyclic groups of order 6 and 30 respectively with generators say A and B respectively.
- Now take the element \(B^5\).The following Lemma describes its order.
- Lemma: If g is the generator of the cyclic group of order n. Then \(g^k\_ has order n/gcd(n,k).(Check !)
- So \(B^5\) has order 6 and hence it is isomorphic to (Z/7Z)*.
- Hence the answer is True.

Bonus Problem:

- Theorem: The group (Z/nZ)* is cyclic if and only if n is \(1, 2, 4, p^k or 2.p^k\), where p is an odd prime and k > 0. This was first proved by Gauss. (Wow!)

Solve and Salvage if Possible.

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## 2 replies on “Multiplicative group from fields: TIFR GS 2018 Part A Problem 17”

Very useful….. Good metod…. Keep on going

Thank you. Keep reading!

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