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

Understand the problem

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

Start with hints

Hint 1

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?

Hint 2

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.}

Hint 3

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.

Hint 4

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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TIFR 2014 Problem 26 Solution -Number of irreducible Polynomials

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Problem

The number of irreducible polynomials of the form (x^2+ax+b) , with (a,b) in the field (\mathbb{F}_7) of 7 elements is:

A. 7

B. 21

C. 35

D. 49

Discussion:

First, what is the number of polynomials of the form (x^2+ax+b) in (\mathbb{F}_7) ? (a) has 7 choices and (b) has 7 choices. So we have a total of (7 \times 7 =49) monic degree 2 polynomials in (\mathbb{F}_7).

How many of these are reducible? If a monic polynomial of degree 2 is reducible then it must break into two factors in the form (p(x)=(x-\alpha)(x-\beta) ).

We want to count the number of polynomials of the form (p(x)=(x-\alpha)(x-\beta)). This will give us the number of reducible polynomials and we will subtract it from the total number to get the number of irreducible.

For the counting purposes, we must be careful not to have repetitions. For (\alpha=0) we can allow (\beta=0,1,..,6). For (\alpha=1) we can allow (\beta=1,...,6) (NOT 0 because that will again give ((x-0)(x-0)) which we have already counted in the (\alpha=0) situation ).

For (\alpha=2) we have (\beta=2,...,6) and so on... for (\alpha=6) we have (\beta=6) only.

So in total, we have (7+6+...+1=\frac{7\times 8}{2} = 28 ) polynomials which are reducible, has degree 2 and is monic.

Therefore, the number of irreducible polynomials are (49-28=21).

HELPDESK

What is this topic: Linear Algebra
What are some of the associated concept: Monic polynomial
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TIFR 2013 problem 9 | Existence of element of order 51

Try this problem from TIFR 2013 problem 9 based on the existence of an element of order 51.

Question: TIFR 2013 problem 9

True/False?

There is an element of order $51$ in $\mathbb{Z_{103}^*}$

Hint: 103 is prime.

Discussion: Since for $p$-prime, $\mathbb{Z_p}$ is a field with addition modulo p and multiplication modulo p as the first and second binary operations respectively, and for a finite field $\mathbb{F}$, $\mathbb{F}^*$ forms a cyclic group with respect to multiplication. Hence $\mathbb{Z_p}^*$ forms a cyclic group.

We have $\mathbb{Z_{103}^*}$, a cyclic group or order $103-1=102$.

For a cyclic group of order $n$, if $d|n$ then there exists an element of order d.

Since $51|102$, there exists an element of order $51$ in $\mathbb{Z_{103}^*}$.

Alternative: By Sylow's first theorem, there exists elements of order 3&17 in $\mathbb{Z_{103}^*}$. ( $|\mathbb{Z_{103}^*}|=102=2*3*17$)

Since the group $\mathbb{Z_{103}^*}$ is abelian, the group and 3,17 are co-prime we get an element of order $3*17=51$ (by simply multiplying the element of order $3$ to the element of order $17$).