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# Multiplicative group from fields: TIFR GS 2018 Part A Problem 17

This problem is a cute and simple application on the Multiplicative group from fields in the abstract algebra section. It appeared in TIFR GS 2018.

# 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
Abstract Algebra
Medium
##### Suggested Book
Dummit and Foote

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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